PLANE
SOLID
AND
GEOMETRY
BY
C.
INSTRUCTOR
OF
MATHEMATICS,
A.
HART
WADLEIGH
HIGH
SCHOOL,
NEW
YORK
CITY
AND
DANIEL
HEAD
OF
OF
DEPARTMENT
THE
WITH
J.
H.
AMERICAN
FELDMAN
ERASMUS
MATHEMATICS,
EDITORIAL
TANNER
PROFESSORS
NEW
D.
YORK
OF
HIGH
SCHOOL,
COOPERATION
MATHEMATICS
IN
CINCINNATI
BOOK
SNYDER
VIRGIL
AND
":"
HALL
CORNELL
UNIVERSITY
":"
CHICAGO
COMPANY
BROOKLYN
OF
H-3
Copyright,
1911,
AMERICAN
Entered
H.-F.
BOOK
at
PLANK
by
COMPANY.
Stationers'
Hall,
AND
W.
v"
1912,
SOLID
P.
GEOMETRY.
I
London.
PREFACE
book
This
in
the
outgrowth
of
mathematics
teaching
has
text
is the
used
been
school
criticisms
the
hundreds
of
the
throughout
is therefore
test, and
The
the
The
the
needed.
and
left
The
in its
book
of
present form
experience,classroom
initiated
of teachers,as
of the
proofs
as
leading features
into
of the
exercises
In
incomplete form.
parts of proofs are thus
in which
the
proof is
an
the
for
easier
the
Solid
left to the
not
of the
in
is such
full.
theorems
student,
Geometry,
student
complete,
the
book
:
subject. Definitions
expressed through
country.
been
given
have
some
the
of theorems
arrangement
Geometry, proofs of
are
of the
as
Associations
the
years
The
teachers, in classes of all
product
some
rapidly
general demand
of
many
schools.
of secondary
varying conditions
proofs have had the benefit of
matics
experienced teachers of mathe-
country.
is
selection
Most
of
combined
following are
given only
are
secondary
criticism.
severe
Tlie student
in
under
The
teaching.
experienceof
an
different
by
many
development, and
stages of
of
In
and
or
are
more
; but
in
as
to meet
the Mathematical
the
Plane
tions
construc-
given in
proofs and
every
case
incompleteness is
stated.
specifically
The
usual
indirect
method
method
of proof
of
is
consistentlyapplied.
The
proving such propositions,for example, as
Arts. 189 and
The
method
415, is confusing to the student.
used
here is convincing and clear.
In choosing exercises,
The
exercises
are
carefully selected.
each of the following groups
has been given due importance :
problems and
(a) Concrete
exercises,including numerical
problems of construction.
ments
(6) So-called practical problems, such as indirect measureof heights and distances
of equal and similar
by means
triangles,drawing to scale as an applicationof similar figures,
from
from physics,
design, etc,
..problems
304026
iv
PREFACE
(c) Traditional
The
exercises
they
without
benefit
of
The
a
more
or
exercises
is
less abstract nature.
pedagogical Comparatively
placedimmediatelyafter
are
the theorems
gether
applications,instead of being grouped toFor
regard to the principlesinvolved in them.
the brighter pupils,however, and for review
are
classes,
long lists of
at the end
of
the
of
arrangement
easy
of which
the
exercises
of each
more
less difficultexercises
or
grouped
are
book.
closed
definitionsof
figures are
The
unique.
student's
natural
conceptionof a plane closed figure,for example, is not
the boundary line only,nor
the plane only,but the whole figure
All
composed of the boundary line and the plane bounded.
definitions of closed figuresinvolve this idea,which
is entirely
consistent
The
with
the
higher mathematics.
the fundamental
of magyiitudes is explicit,
principlesbeing definitelyassumed
(Art. 336, proof
numerical
treatment
a
Appendix, Art. 595). This novel procedure furnishes
of dealing with
logical,as well as a teachable, method
in
incommensurables.
The
of a rectangleis
area
thereby obtaining its
the
order
same
of theorems
the
in this connection
is valuable.
introducingthe
Proofs of
the
triangle.
volume
The
A
of
a
as
correlation
similar
permits
is used
with
is
method
in the
arithmetic
employed
parallelopiped.
the
and
accurate
exceptionally
historical notes
many
The
superposition theorems
will be found
theorems
corollaries
and
method
This
measure-number.
parallelogramand
for
by actuallymeasuring it,
introduced
ivill add
and
lifeand
line
concurrent
complete.
interest
to
the
work.
The
throughout the book, the
formulas
of Plane Geometry, and the collection of
Solid Geometry, it is hoped, will be found
helpful
carefullyarranged
collection
formulas
to teacher
of
of
and
Argument
arrangement
the
alike.
student
and
reasons
gives a
careless
leads
impossible,
summaries
arranged
are
definite
omission
to accurate
labor of reading papers.
model
of the
for
reasons
parallelform. This
ders
proving exercises,ren-
in
in
a
demonstration
thinking,and greatlylightensthe
PREFACE
Every
construction
This
method
figure
contains
v
all
necessary
construction
keeps constantlybefore the student
a
model
for his construction
work, and distinguishesbetween
a
figurefor a construction and a figurefor a theorem.
TJie mechanical
arrangement is such as to give the student every
possibleaid in comprehending the subjectmatter.
of the specialfeatures of the Solid
The following are
some
Geometry :
The vital relation of the Solid Geometry to the Plane
Geometry
is emphasized at every point. (See Arts. 703,786, 794, 813, 853,
924, 951, 955, 961, etc.)
The
student
is given every possible aid in forming his early
in Solid
concepts. In the early work
Geometry, the
space
student
in fullycomprehending
experiences difficulty
average
relations,that is, in seeing geometric figuresin space.
space
The student is aided in overcoming this difficulty
duction
by the introof many
easy and practicalquestions and exercises,as
his figures. (See " 605.)
well as by being encouraged to make
further
As
aid in this direction, reproductions of models
a
in a group (p.302) and
shown
made
are
by students themselves
at various
points throughout Book VI.
is constantly
Tlie student's knowledge of the thingsabout him
draivn
Especiallyis this true of the work on the sphere,
upon.
the student's knowledge of mathematical
where
geography has
and the relations
been appealed to in making clear the terms
with the sphere.
of figuresconnected
the demonstrations
in
The same
logicalrigor that characterizes
the Plane
Geometry is used throughout the Solid.
The
treatment
of the polyhedral angle (p. 336), of the prism
(p. 345), and of the pyramid (p.350) is similar to that of the
with
the
This
is in accordance
cylinder and of the cone.
Associations
recommendations
of the leading Mathematical
throughout the country.
is due to many
The gratefulacknowledgment of the authors
A.
Grace
friends for helpful suggestions; especiallyto Miss
York; to Mr.
Bruce, of the Wadleigh High School, New
Edward
B. Parsons, of the Boys' High School, Brooklyn ; and
of Cornell University.
to Professor
McMahon,
lines.
CONTENTS
PLANE
GEOMETRY
PAGE
SYMBOLS
viii
ABBREVIATIONS
AND
1
INTRODUCTION
BOOK
Polygons.
21
Proof
for
Necessity
19
FIGURES
RECTILINEAR
I.
22
Triangles
25
Superposition
of Distances
Measurement
by
of
Means
Triangles
32
...
42
Loci
Parallel
65
Lines
Construction
of
for
Directions
Miscellaneous
BOOK
Theorems
100
Triangles
106
Line
Concurrent
THE
II.
84
Parallelograms
Quadrilaterals.
of
Solution
the
110
Exercises
Exercises
Ill
CIRCLE
113
131
Circles
Two
.
133
Measurement
BOOK
156
Exercises
Miscellaneous
FIGURES
SIMILAR
AND
PROPORTION
III.
161
.
.
Similar
Drawing
Polygons
176
Scale
188
to
Miscellaneous
BOOK
AREAS
IV.
Transformation
Miscellaneous
206
Exercises
of
223
Figures
239
Exercises
.
BOOK
V.
209
POLYGONS
OF
.
....
.
MEASUREMENT
POLYGONS.
REGULAR
.
THE
OF
245
CIRCLE
Measurement
Miscellaneous
of the
and
Circumference
of
the
Circle
258
.
.
274
Exercises
vi
vii
CONTENTS
PAGE
OF
FORMULAS
TO
PLANE
and
Minima
APPENDIX
Maxima
PLANE
Variables
and
GEOMETRY
VI.
284
Theorems
Theorems
LINES,
.282
.
.
284
291
294-298
........
GEOMETRY
SOLID
BOOK
.
GEOMETRY
Limits.
Miscellaneous
.
AND
PLANES,
ANGLES
SPACE
IN
299
.
*
Planes
and
Lines
Dihedral
Angles
301
322
*
.
POLYHEDRONS
VII.
.
.
Polyhedral Angles
BOOK
.
.
.
.
.336
..
.343
.
...
Prisms
345
Pyramids
350
of the Prism
Mensuration
and
354
Pyramid
Areas
354
Volumes
.
Miscellaneous
BOOK
VIII.
358
.
Exercises
381
CYLINDERS
AND
383
CONES
383
Cylinders
Cones
of the
Mensuration
Cylinder and
Cone
392
Areas
392
Volumes
406
Miscellaneous
IX.
BOOK
388
"
Lines
417
SPHERE
THE
and
414
Exercises
Planes
Tangent
to
a
424
Sphere
429
SphericalPolygons
Mensuration
of the
Areas
.
444
Sphere
.
444
,
Volumes
457
Miscellaneous
FORMULAS
OF
APPENDIX
TO
Exercises
SOLID
SOLID
on
Solid
GEOMETRY
GEOMETRY
Geometry
467
....
.
.
.
.
.471
474
SphericalSegments
474
The
475
Prismatoid
Similar
.INDEX
Polyhedrons
477
481
AND
SYMBOLS
ABBREVIATIONS
equals, equal to, is equal
=
=fc does
equal.
not
"
greater
"
less
than, is greater
than.
is less than.
than,
equivalent
equivalent,
=0=
to.
to, is
lent
equiva-
to.
to, is similar
similar, similar
~
S^ is measured
JL
by.
perpendicular
perpendicular,
perpendicular
Js
to.
to,
is
to.
perpendiculars.
|| parallel, parallel to, is parallel
to.
||s parallels.
and
.
"7,
.
so
(sign of continuation).
on
.
v
since.
.-,
therefore.
/-n
arc
OD
;
AB,
AB.
arc
parallelogram,
parallelograms.
O, (D circle, circles.
Z, A
A,
A
angle, angles.
triangle, triangles.
q.e.d.
Quod
erat
demonstrandum,
q.e.f.
Quod
erat
faciendum,
The
signs +,
"
,
X
+
have
which
which
the
same
,
viii
was
was
to
to
be
proved.
be done.
meanings
as
in
algebra.
GEOMETRY
PLANE
INTRODUCTION
1.
we
The
Subject
shall
continue
will
work
be
geometric
the
see
the
with
he
which
Exercise.
color
?
What
do
the
from
apart
it
give
you
What
is it that
is
this
surface
?
intersections
thick
where
that
If you
will
move
a
along
itself.
generate
surface
so
?
so
will
What
that
Can
that
it will
you
move
block
of
wood.
may
be
led
he
to
geometric figures
to the
block
the
it will
from
has
?
name
of
edges
not
generate
so
is and
?
the
block
a
surface
that
do
wide
the
are
Do
lines.
are
how
?
distinguish from
so
point.
a
a
How
Where
?
will it generate
generate
?
intersections
these
It is
generate
They
What
size.
How
?
?
one
what
space,
point
there
are
this
where
the
solid.
the
space
block
the
name
wide
How
and
Shape
?
surrounding
are
their
is
not
?
jj
possessed
block
of
properties
weight
,
surfaces
through
it
Has
box).
solid.
solid
What
say
chalk
the
geometric
?
a
this
intersections
a
the
would
name
is its
of
with
as
Imagine
?
many
you
a
or
(" 13)
physical
imagine
the
?
What
surfaces
the
A
where
block
the
What
edge
is ?
move
What
How
Can
?
long
one
?
many
?
wood
you
long
intersect
how
?
how
?
lines
these
How
A
geometric
?
ball
called
are
?
Can
it that
this
separates
intersect
they
an
has
(or
properties of
of
?
space
properties
thick
block
occupied
this
to
of wood
These
of the
removed.
which
space
the
familiar
main
our
familiar.
solid
a
think
you
block
such
already
solids
physical
size ?
?
shape
defined
following exercise,
block
a
algebra,
be
a
as
although
geometry,
and
later
is
such
become
at
call
we
Can
solid.
solid
?
taste
such
must
Look
will
the
of
of
relation
what
him,
about
study
arithmetic
student
The
figures.
careful
a
of
In
Geometry.
of
use
study
a
physical objects
By
of
Matter
all
solid
of
?
generate
it will
What
?
not
it
as
them
?
Can
Yes,
by
moving
?
Can
will
generate
you
the
a
?
moves
you
it
move
corners
line ?
PLANE
FOUR
FUNDAMENTAL
The
2.
GEOMETRY
GEOMETRIC
in which
CONCEPTS
and
live,although boundless
unlimited
in extent, may
be thought of as divided
into parts.
A physicalsolid occupies a limited portion of
The
space.
portion of space occupied by a physical solid is called a
space
solid.
geometric
A
3.
geometric solid has
It may
also be divided
called
surface
breadth,but
The
A
of
a
of
part of
no
divided
a
FOUR
As
we
have
small
it
solid is
dot
of
line.
a
If
it build
point is
a
will be
moves
a
be divided
length
into
and
parts.
It has
length only. It
boundary or extremity
It has
IN
up
with
a
may
to the
allowed
to
length, nor
parts. It is
ORDER
solid
consider
independentlyof
dently,
point indepen-
solid.
sharp pencilon
move
into
REVERSE
geometric
we
neither
be divided
It cannot
point,so
made
has
It
point.
represents approximatelya geometric
7.
a
line.
parts. The
considered
from
a
surface.
CONCEPTS
surface,line,and
and
a
into
point is no part
breadth,nor thickness.
positiononly.
THE
also
is called
part of
no
A
A
thickness.
boundary of
solid.
a
It may
surface
a
line is called
6.
parts. The
thickness.
line is
also be
may
is
no
boundary
5.
into
length,breadth, and
surface.
a
A
4.
we
a
sheet
of paper
path
in which
point.
in space, the
line.
piece of fine wire, or. a line drawn on paper with a sharp
ever
pencil,represents approximately a geometric line. This, howA
fine it may
be, has
some
thickness
and
is not
therefore
an
ideal,or geometric,line.
8.
If
will be
a
a
line is allowed
surface.
to move
in space, its
path in general
INTRODUCTION
9. If
surface
a
general will
dimensions
dimensions
two
a
point has
11.
these
The
;
a
geometric
A
surface
is that
adjoining solid
A
line
A
point is
has
limited
a
which
path
in
three
have
to
dimension
working definitions
as
;
dimensions.
no
of
has
solid
a
space.
separates it from
or
the
surrounding space.
length only.
positiononly.
AND
ASSUMPTIONS
primary objectof elementary geometry
is to determine,
by a definite process of reasoning that will be introduced
developed later,the propertiesof geometric figures. In
and
all
of this
logicalarguments
fundamental
upon
principlesare
these
as
a
foundation
The
assumptions here
and
axioms
the
13.
solid,or
14.
A
Def.
the book
geometric
combination
a
Def.
mentioned
postulates. These,
given throughout
Geometry
in
kind, just as
agreed upon
well
figure is
of any
outset, and
a
the
as
In
mentary
ele-
called
into two
classes,
will
definitions,
for them
arises.
point,line,surface,or
all of these.
or
is the
the
is built.
divided
occasion
as
debate,certain
principlesare
are
as
a
at
argument
geometry these fundamental
and assumptions.
definitions
be
one
or
portion of
bounds
from
or
is that which
The
used
is said
and
extent
therefore
be
DEFINITIONS
12.
its
concepts :
is
solid
space,
is said to have
so
extent
onefold
following may
four fundamental
and
twofold
line has
and
in
move
extent
has
extent
no
A
an
threefold
surface
a
;
to
solid.
geometric
a
solid has
A
10.
be
is allowed
3
science
which
treats
of
the
propertiesof geometric figures.
15.
A
be
defined
the
assumption of
the possibility
of performing a certain geometric operation.
Before
it will be necessary to
giving the next definition,
Def.
introduce
a
postulate
nostulate.
may
as
4
PLANE
16.
Transference
moved
from
one
Def.
Two
GEOMETRY
Any
postulate.
another
position to
be
geometric figure may
without
change of
size
or
shape.
17.
geometric figures are
either is
when
placed
point of the
some
upon
18.
Two
Def.
the
upon
said
other,each
to
if,
coincide
point
of
lies
one
other.
geometric figuresare
equal if
they
be
can
to coincide.
made
19.
The
Def.
shall coincide
that the two
is
This
of
process
placing one
another
so
is called superposition.
actual
imaginary operation,no
an
figureupon
taking
movement
place.
LINES
line
line is
A
20.
AB.
also
by
a
somewhere
21.
or
It may
small
Straight
that
line
a
a
capitalletters,as
two
be
designated
letter placed
the
on
learned
usuallydesignated by
a.
" 7
we
fine wire
piece of
on
"
,
In
Lines.
drawn
line
line,as
.
Fig.
of
sheet
a
1.
represented approximatelya geometric line. So also a
be represented approximatelyby
geometric straightline may
two
a
string stretched taut between
points,or by the line
flat
made
on
a
by placing a ruler (also called a straightedge)
surface and drawing a sharp pencilalong its edge.
paper
22.
edge
of it and
were
you
test
the
piece,so
length of
piece
straight?
cut
that
the
out
gardener
a
does
How
straight,but
slipit along the
entire
Would
?
not
this
the
bed
could
How
wire
a
flower
a
?
rows
of
of
does
How
Questions.
wire
such
shape
that
so
wire
straightnessof
that
a
you
it would
wire
?
could
always
changed places,would
its ends
?
the
get his plants set
straightness of
the
of
he
test
If
fit under
you
these
turned
various
out
Can
cut
fit ?
in straight
think
you
out
piece
a
If you
it stillfit
it over,
conditions
the
would
versed
re-
along
it fit ?
if the
wire
INTRODUCTION
23.
Def.
of it is
placed with
placedwill
Thus, if
CD, is
of
straight
A
is
AB
a
any other
its ends
in
part
with
AB,
will lie in
from
line lie
(" 23)?
Draw
Show
P.
that
AB
word
The
line,
straightline.
A
straightline
Can
AB.
be drawn
may
will every
that
follows
between
be drawn
can
draw
you
If so, where
It then
and
9
5
line.
right
a
mean
second
a
pointof
the
:
points; i.e.a
two
by two points.
straightlines
two
as
other.
any
line
Only one straight
line is determined
straight
26.
to
to B ?
A
S
A
also
straight line
a
straightline
second
portionso
FlG- 2-
postulate.
point to
one
Draw
25.
entire
AB.
is called
line
Straight
any
line,the
portion
any
line,howrever it may be applied.
straightline,and if any portion of AB,
is understood
unqualified,
from
in the
its ends
that,if
lie in the
pointof CD
straightline
24.
line such
a
placed on
AB,
every
A
is
line
AB
cannot
CD
and
CD
have
a
in point
intersecting
second point in common
(" 23). It then follows that:
Two
lines can
have only one
intersecting
straight
i.e.two intersecting
lines determine
straight
27.
segment,
CD, and
28.
line
limited
simply a
or
DB
Two
A
collinear
curved
curve) is a line no
which
is straight,
as
30.
A
Def.
line made
up
broken
4
a
lie in the
same
segments.
(or
portionof
line
GH.
line
of different
straightlines,
as
*
point.
a
line segments which
said to be
Def.
mon;
com-
straightline is called a line
Thus, in Fig.2, AC,
segment.
portionof
line, or
a
in
line segments.
are
Def.
are
29.
A
Def.
point
KL.
is
a
sive
succes-
Fig.
3.
straight
6
TLANE
31.
Use
in the
Only two instruments
of Instruments.
constructions
edge for drawing
the
GEOMETRY
a
for
compasses,
plane geometry: the ruler or straightstraightline,alreadyspoken of in " 21 ; and
constructingcircles or arcs of circles,
and
for
line segments
transferring
Thus to add two lines,
and
AB
as
CD, draw,
with a ruler,a straight
line OX.
Place one
leg
of the
and
Next
place
and
0
CD
+
Ex.
Can
straightline
a
curved
move
line
move
broken
line
move
three
lines
as
Draw
4.
of
EF;
Ex.
What
ON.
so
that
in
space
its
is the remainder?
path will
be
not
surface
a
?
its
path will
its
path
that
so
not
be
a
will not
be
a
?
the
in space
AB,
and
CD
and
EF;
and
EF.
Add
is the
6.
Jhree times
and
CD,
that
so
CD
;
(c)
the
(6) the
the
Draw
as
a
long
CD
the
difference
for
line twice
AB.
^
; of AB
difference
for
be-
E
between
FlG
5
AB
between
results obtained
as
and
C
difference
result obtained
CD,
EF.
(")
:
AB,
and
of AB
sum
Construct
5.
AB
Ex.
from
AB
a
a
Construct
tween
ON.
=
?
Ex.
and
a
Then
If so, how
3.
surface
EF.
In
to subtract
Can
?
Ex.
at
?
2.
surface
leg
MN
Can
If so, how
B.
CD.
how
1.
Ex.
another.
lay off
0M+
"
Show
positionto
one
A
at
AB.
manner
equal to
MN
from
off segment
cut
similar
sum
one
equal to
OM
AB
at
compasses
other
the
permitted
are
of
(a)
and
(6)
and
see
whether
the
(c).
as
long
as
AB
(the sum
of AB
and
AB)
;
INTRODUCTION
SURFACES
32.
It is well
Surface.
Plane
known
surfaces
straightedgeis applied to
that
to
test
the
whether
carpenter's
they are
straightedgeis placed
the surface is called a plane.
the surface,it always fits,
on
should
less
doubtshould use a powerful magnifier,we
if we
Now
discover that in certain places the straightedgedid not
placed. A sheet of
exactly fit the surface on which it was
flat and
fine
Can
a
plane.
How
of
a
surface
such
would
the ruler
fit in
but
(a great many)
of
think
that
such
and
out
34.
ruler ?
the
the rest
be
but
cut
that
so
it fit if turned
points are
of unlimited
whatever
Fig.
6.
two
taken, a straightline joiningthem
will
lie
the surface.
wholly in
35.
that
(or
surface
plane
surface
a
such
extent
of its
A
Def.
plane)is
Def.
is
A
curved
A
plane figure is
surface
a
surface
no
which
portion of
plane.
36.
Def.
in
pointslie
37.
Ex.
7.
plane.
one
A
Def.
of which
rectilinear
Plane
geometric figureall of whose
treats of plane figures.
Geometry
a
figure is
a
plane figureall
the
lines
straight lines.
are
How
can
a
plane
move
in space
so
that its path will not
be
solid ?
Ex.
be
apply
(insideout) ?
over
a
of it could
Would
fit ?
you
ruler to
a
Can
plane
not
slippedalong
must
with
that
in all ?
surface
piece
a
it would
is
a
blackboard
the
times
many
ideal.
positions
some
not
of
surface
the
Test
it is
think
you
you
nearlyapproachesthe
Questions.
whether
the
where
matter
plateglassmore
33.
see
If, no
even.
a
8.
solid ?
Can
a
curved
If so, how
surface
?
move
in space
so
that
its
path
will not
GEOMETRY
PLANE
ANGLES
38.
diverge from
point is the vertex
lines which
The
figure formed
point.
angle is the
An
Def.
a
of the
by
the
angle and
straight
two
lines
its
are
sides.
39.
An
be designated
angle may
it,as angle 1 and
angle
2 in
by
number
a
Fig. 7, and
angle
L
placed
3 in
within
Fig. 8.
Or
H
M
Fig. 9.
three
letters may
be
used, one
vertex, the last being read between
angle 1 may be read angle ABC,
angle is often designated also by
other angle has the same
when
no
40.
Revolution
plane,about
a
uouslyfrom
one
41.
and
the
in
and
positionFE
42.
vertex
we
and
the
its vertex,
singleletter at
vertex, as angle F
straightline
when
Fig. 8.
in
revolve
may
it does
another, it passes
revolve
in
contin-
and
only
angle
may
once
a
Two
Def.
and
a
magnitude of
revolved
sides
about
imagine
to have
and
angles are
common
1 and
angle KML
,
once
position.
may
then
the
at
one
an
at first collinear,
were
point common
a
to the
first to have
FG
revolved
be
about
F
as
been
a
in
pivot
positionFG.
Fig. 7, angle
JIMK
Fig. 8.
and
two; thus in Fig. 7,
An
angle 2, angle CBD.
that its two
has
side
the other
pivot, and
of the
notion
of them
one
Thus
to the
a
by imagining
that
two.
as
position to
clear
A
obtained
point
intermediate
through every
A
postulate.
each
on
adjacent
side which
if
they
lies between
angle 2 are adjacent; also
are
adjacent.
have
them
in
a
common
; thus
in
Fig. 9, angle
10
PLANE
46.
If
Def.
each
47.
An
Def.
lines meet, hut
two
other, they
said
are
to
oblique to each
is
an
right angle ;
a
perpendicular to
not
are
be
angle
acute
that is less than
GEOMETRY
as
other.
angle
angle 1,
Fig. 13.
48.
An
Def.
obtuse
that
is greater than
than
two
49.
angle
an
less
rightangle and
a
right angles;
A
is
angle
angle 2, Fig. 14.
as
is
angle
is greater than two right angles and
than
four right angles; as
angle 2,
that
less
Def.
reflex
angle
an
Fig. 15.
50.
form
two
and
of
the reflex
as
BA
as
formed
by
from
and
BC,
formed
9.
Ex.
Make
(b)
and
of
What
position
Angle 2 may
line counter-clockwise
a
should
be
read
Fig. 16, if angle
angles are they ?
with
statement
angle
be
from
thought of as
the positionBC
CBA.
reflex angles
are
sometimes
1
equals angle
regard
D
lines
to the
7?
angles
a
does
3
angle
and
Ex.
ABC.
In
of
a
(eZ)Make
AB
the
to
CD.
If
(c)
kind
and
thought
line counterclockwise
a
angles.
(a)
kind
of
position BA
read
angle
of
be
1 may
Acute, obtuse, and
oblique
what
Fig. 15, always
acute
angle 1
revolution
revolution
the
Def.
called
AB
be
position BA,
51.
2,
the
should
by
to the
the
the
the
Angle
angle 2.
from
diverging
and
BC,
positive angles, as
point,
same
lines
Two
Note.
they
are
equal angle 4,
not
what
?
with
statement
regard
to the
lines
Fig.
CE.
10.
A
kind
of
blackboard
plumb
angles
? with
a
line is
does
line
on
suspended
it make
with
the blackboard
from
a
the
top of the blackboard.
horizontal
neither
16.
line drawn
horizontal
nor
on
the
vertical ?
11
INTRODUCTION
the
may
(a)
hand
hour
acute
an
12.
Ex.
be
angle ?
Draw:
hand
minute
the
Suppose
11.
Ex.
the
that
so
of
clock
a
hands
two
is at
twelve.
with
make
obtuse
("")a
(a)
pair of adjacent angles; (6)
a
an
other
:
angle ?
rightangle ?
(c)
Where
each
a
of
pair
non-
adjacent angles.
Draw
13.
Ex.
angle ; (")
each
each
is
the
other
and
is acute
adjacent angles such
two
rightangle ; (c)
right; (e) one is acute
a
Fig. 17, angle 1 -f angle 2
? angle BAD
3 + angle 4
+ angle DAF
? angle 2 + angle DAF
DAF
angle 3
?
? angle 4 + angle BAE
=
angle 1
In
14.
Ex.
=
Fig.
(c)
Draw
16.
(a)
:
a
a
pairs
an
and
angle ; (e?)one
other
the
acute
an
obtuse.
angle
?
angle
angle 4
?
=
"
adjacent angles
of
in
A
Def.
is divided
53.
The
bisectingAB
side ;
common
side.
common
bisector
and
if it
FlG'
of
an
angle is the
line which
17'
divides
equal angles.*
line
AB, neither horizontal
and
not
perpendicular to AB
not perpendicular to it.
a
line
a
a
a
equal parts.
two
Draw
17.
(")
;
line is said to be bisected
angle into
freehand
and
into two
Def.
Ex.
vertex
vertex
that
non-adjacent angles
two
common
common
52.
the
six
obtuse
is
each
17.
Ex.
have
Name
15.
(a)
:
=
"
Ex.
is
=
=
"
that
vertical,
nor
bisectingit :
(a)
(6)
Draw
a
line
ASSUMPTIONS
1.
54.
the
Things equal to
equal to each
same
thing,or
to
equal things,are
other.
2.
If equals are
3.
If equals
are
added
to
equals,the
subtracted
from
sums
equal.
are
equals, the
remainders
are
equal.
4.
the
If equals are
If equals are
in the
same
proofthat
every
unequal
'
*
The
to
unequals, the
sums
are
unequal in
order.
same
5.
added
subtracted
from unequals, the
remainders
are
order.
angle has
but
one
bisector
will be
found
in the
Appendix, " 599,
12
PLANE
6.
If unequals are
in the
unequal
subtracted
(a) If equals are
equal; (6) if unequals
in the
8.
(a)If equalsare
multipliedby equals,the products are
are
multipliedby equals,the products are
divided
by equals,the quotientsare equal;
by equals,the quotientsare unequal
divided
If unequals
are
added
unequals,the
to
greaterto the greater,the sums
.10. If three magnitudes of the
the
less to the less and
unequal
are
kind
same
in the
are
the
firstis greater than the second and the second
third,then the firstis greater than the third.
11.
The
whole
is
equal to
12.
The
whole
is
greater than
13.
Like powers
equal numbers
14.
from
16.
Straight
point to
one
any
Straight
line
Revolution
plane, about
from
a
an
A
axiom
Ex.
18.
numbers
Ex.
are
like roots
equal,and
of
Any geometric figure may
another
without
change of size
to
19.
A
straightline
(See " 24.)
other.
any
point
A
H.
postulate
be
or
line
segment
A
as
straightline
pivot,and
a
when
be drawn
be prolonged
may
revolve
revolve
once
and
in
a
ously
continu-
only once
position. (See " 40.)
intermediate
ssumptions 1-13
be
may
it does
position to another,it passes
may
may
either end.
postulate.
one
throughevery
55.
its parts.
of its parts.
any
postulate I.
line
at
indefinitely
17.
the
greater than
(See " 16.)
shape.
15.
related that
so
postulate.
position
one
of all
sum
order.
same
equal.
are
from
the
of equal numbers
Transference
moved
are
order.
same
9.
remainders
order.
same
(b) if unequals are
in the
from equals,the
order.
reverse
7.
unequal
GEOMETRY
are
denned
usuallycalled axioms.
a
as
Illustrate the first five
statement
whose
assumptions above
That
truth
is
is,
sumed.*
as-
cal
by using arithmeti-
only.
Illustrate
the next
five
by using generalnumbers
only
*
See
Appendix, " 600.
(letters)
13
INTRODUCTION
DEMONSTRATIONS
It has
been
("12) that
stated
the fundamental
principles
cal
agreed upon at the outset as forming the basis of the logiaxioms, and
arguments in geometry are called definitions,
56.
postulates. Every
of
statement
by
must
is called
The
proof
a
has
establish
a
or
thus
or
been
process
established,it
in turn
be
a
formed,
per-
depend
reasoning
to
of
the truth
After
demonstration.
of
be
may
mentioned
are
divided
a
ment
state-
used
to
into two
classes,
is
required
problems.
A
Def.
is
theorem
a
statement
whose
truth
to be
a
to
reasoning be shown
of
it is
truths.
new
and
57.
something
principles. This
propositionshere
theorems
of
statement
a
process
fundamental
these
upon
truth
a
whether
propositionadvanced,
new
For example,
If two anglesof
proved or demonstrated.
are
equal,the sides oppositeare equal" is a theorem.
triangle
"
There
the
two
are
conditional
proved.
parts
each
Write
20.
of the
(a) If
out
:
conclusion,
the
or
hypothesis,
the
part
to
or
be
angle
of a trijust quoted, If two a7igles
the sides oppositeare
hypothesis; and
"
"
equal " is the conclusion.
Ex.
the
theorem
equal " is the
are
theorem
every
part ; and
the
In
to
.
carefullythe hypothesis
and
the
conclusion
of
following :
times,you will be rewarded.
learn geometry.
(6) If you try to memorize
proofs,you will never
your
(c) You must suffer if you disobey a law of nature.
(d) Things equal to the same
thing are equal to each other.
(e) All rightangles are equal.
58.
you
do your
Def.
A
from
another
theorem, must
59.
A
at all
duty
corollary is
be
truth.
a
statement
Its
of
a
truth
duced
easilyde-
correctness, like that
of
a
proved.
in
general,is a question to be solved.
As applied to geometry, problems are
of two
kinds,namely,
of construction
and problems of computation.
problems
Def.
problem,
14
PLANE
the
From
60.
definition of
a
Cor.
revolution
postulate(" 40) and
perpendicular(" 45) we may deduce
a
corollaries
61.
the
from
ing
the follow-
:
At
I.
perpendicular
line
Given
GEOMETRY
in
every
to the
point
and
point
AB
a
straight
line there exists
line.
in
C
AB.
To
at a
AB
exists
that there
prove
JL to
a
en.
as
c
CE
Let
Z 1 "
(Fig.18)
O,
at
is revolved
if CE
Then
Z 2.
AB
meet
about
position CA,
pivot toward
decrease
will continuously
("40)as
as
a
in which
CD,
CD"AB
the
A
two
that
so
Z
formed
Fig.
18.
C
1
will
.-.
there
with
B
continuously increase
be
must
AB
are
and
Z
2
position of CE,
one
In
equal.
this
position
line
there
("45).
62.
Cor.
only
exists
Given
point in a straight
perpendicular to the line.
At
II.
one
A.
CD
every
at
AB
C, i.e. Z
BCD
=
J)
E
/.DCA.
i
\
To
at
the
CD
prove
only
"
JL to
AB
\
\
C.
Let
CE
other
(Fig
CD.
than
(M*
and
and
follows
At
ZECA
CE
is not
and
point
perpendicular
A
line
fall between
to
62
J.i?
may
C
A
CD
B
Fig.
19.
and
EGA
i.e."
"ZDCA
"
C
from
(" 54, 12).
.-.
A BCE
are
not
(" 45).
be
combined
in
one
statement
as
:
every
*
any
ZBCE"ZBCD;
And
""61
63.
be
C#
Let
Then
ZDCA.
equal
19)
similar
in
a
straightline
there exists
one
and
only one
to the line.
proof may
be
given
if
we
suppose
CE
to
fall between
CB
and
CD.
15
INTRODUCTION
64.
Cor.
All
m.
right angles
equal,
are
D
B
Fig.
AAOB
Given
To
Place
Z
that
PC
so
AAOB
prove
OA
CPD
Cor.
line,the
To
of the
str. line CD
C, forming A
at
CE
Then
be
Z BCD
side
one
1 rt. Z
=
\rt.Z
equals
two
all the A
T/^e
a
about
Z4
+ Z
The
.
sum
=
5 +
DC
coincide
A6
of
sum
point 0
-ZDCE.
B
ZDCE.
+
Fig.
all the
angles about
passing through
line
on
+ Z3
21.
point on
that
point
a
6,
one
+
2 rt. A.
of the six A
are
2 rt. A.
"
CA.
=
and
E
A.
2 rt. A.
straight
Z1+Z2
prove
A
two
right angles.
side of str. line
To
A
1, 2, 3, 4, 5, and
A
Given
and
PD
C
V.
of
.-.
the
1 rt. Z ;
=
+ Z 2XL4
Cor.
Then
point O, and
OB, both being
fall upon
(" 63).
Z BCD
Z DC
Z BCD
C
A
=
Again
66.
at
DC
+ Z DCE
.-.
/.
and
Z
shall
str. line
meeting
+
to AB
"
two
(" 62).
P
straight line meets another straight
adjacent angles is two right angles.
BCD
Z BCD
prove
Let
one
point
OA.
with
collinear
If
IV.
sum
Given
AB
must
that
so
shall be collinear
at
65.
CPD.
Z A OB
be
rt. A.
two
CPD, any
A
"
upon
0,
equal ("18).
_L to
Z
and
20.
(Fig.22)is equal to
ZAOB
+
Z BOC.
16
PLANE
67.
of all the angles
point equals four right angles.
Cor.
about
a
Prolong
Cor.
The
VI.
GEOMETRY
sum
of the lines
one
through
the vertex
and
apply
V.
68.
Two
Def.
of two
such
is
sum
called complementary
angle are
Either
angles whose
right
one
angles,as angles 1
angles is
said
Fig.
23.
2, Fig.25.
and
to be the complement
of the
other.
69.
Two
Def.
angles whose
called supplementary
of two
is two
sum
angles,as angles1
and
right angles are
Either
2, Fig. 26.
such
of the other.
angles is said to be the supplement
An
of two
angle that is equal to the sum
right angles
sometimes
called a straight angle, as angle ABC, Fig. 26.
70.
Two
Def.
if the sides of each
vertical
of
the
sides
thus, in Fig. 24, angles 1
angles,and angles 3
vertical angles.
71.
said
angles are
An
Def.
of
a
contains
angle
of
the
2
and
4
and
one
other;
are
is
degree
In
=
Ex.
wise
like-
Fig.
an
22.
35" ? of a0 ?
tieth
nine-
one
angle of
How
of
a
right angle.
in angle
the complement
there
in
23.
"
""
40" ?
many
degrees are
$ rightangles ?
of k
there
in the
complement
of 20" ?
rightangles?
Fig.26, angle 1 + angle 2, or angle
40", C
ABC, equals 2 right angles. If angle 1
how
degrees are there in angle 2 ? How
many
degrees,then, are there in the supplement of an
many
Ex.
24.
rightangle. A rightangle,therefore,
90 angle degrees.
21.
of
tical
ver-
are
is
Fig. 25, angle ABC
If angle 1
40", how
degrees are
many
2 ?
How
degrees,then, are there
many
Ex.
be
to
the prolongations
are
of
is
In
i.
B
=
FlG-
*""
angle of
40"
of
18
PLANE
36.
Ex.
Can
(a)
two
GEOMETRY
angles
be
and
supplementary
adjacent?
not
Illustrate.
(6)
Can
two
angles
be
adjacent
(c)
Can
two
angles
be
both
(d)
Can
two
(e)
Can
two
(/)
Can
angles
be
angles
be
and
supplementary
both
angles
two
supplementary
not
supplementary
neither
be
and
both
?
adjacent
?
adjacent
nor
complementary
Illustrate.
?
and
?
adjacent
complementary
and
Illustrate.
Illustrate.
Illustrate.
?
supplementary
Illustrate.
the
From
73.
angles
follows
Cor.
angles
three
Cor.
To
prove
If
BD
If
III.
ABC
the
the
not
other
sides
Z
+
BD
is
some
of the
corollaries
as
same
angle
or
of equal
same
angle
or
of equal
line
CBD
2
=
rt. A.
of
of
prolongation
B,
AB.
AB,
BE,
as
D
must
B
prolongation.
Fig.
Then
ZABC+Z
Also
ZABC
CBD
+
Z
impossible
of AB
CBD
=
(" 54, 12)
leads, by
Z
.-.
CBE.
BD
correct
Hence
is the
-
CBE
.-.ZCBD
is
tary,
supplemen-
are
eollinear.
are
prolongation
from
angles
adjacent
two
exterior
Z
Given
Z
mentary
supple-
equal.
their
that
additional
of the
Supplements
II.
are
76.
This
and
equal.
Cor.
angles
its
complementary
deduced
Complements
I.
are
75.
be
be
may
of
:
74.
then
definitions
=
=
; i.e. the
2 rt. A
2
rt.
ZCBE
of AB.
hyp.).
("75).
to
supposition
prolongation
29.
(" 65).
supposition
reasoning,
this
A
(By
A
the
that
BE
impossible
itself is false.
is the
longation
pro-
conclusion
I
BOOK
FIGURES
RECTILINEAR
I.
Proposition
77.
If
Theorem
straight lines intersect, the vertical
two
angles
are
equal.
2 and
3, also
and
To
intersecting
two
Given
Z
prove
str.
vertical
the
lines,forming
A,
1
4.
1
Z3
=
Z2
and
=
Z4.
Argument
Reasons
.
Zl
+
Z2
A.
2rt.
=
1.
If
line
str.
one
line, the
str.
of
the
Z3
+
.\Z1
Z2
+
A.
2rt.
=
Z2
Z3
=
+
Z2.
2.
Same
3.
Things
reason
Z3.
4.
If
equal
Likewise
Z
2
Z
=
5.
4.
Cor.
adjacent
equal,
and
If
angles
each
two
straight
thus
angle
lines
equals,
formed
is
a
are
equal,
right angle.
19
same
to
each
ders
remain-
the
" 54, 3.
equal.
to
1, 2, 3,
4.
intersect
2
subtracted
are
steps similar
and
Q.E.D.
78.
By
is
" 54, 1.
are
5.
the
to
equal
are
equals
from
sum
1.
as
other.
=
other
an-
" 65.
thing
.Z1
A
adj.
two
rt. A.
meets
so
all
that
the
any
angles
two
are
20
PLANE
GEOMETRY
If three
straightlines intersect at a common
angles, no two of which are adjacent, as
accompanying diagram, is equal to two
Ex.
37.
of any
the
point,the
three
1, 3,
and
sum
5,
in
rightangles.
In
38.
Ex.
(a) Angle
(6) Angle
other
=
7
=
"
angle 4
show
that
:
angle 7.
angle 6.
angle
"
+
angle 6
The
line which
bisects
The
bisectors
3
twice
=
2
=
one
rightangles.
of two
vertical
angles bisects
the
also.
Ex.
an
angle 3
angle 4
+ angle 4
2 +
(d) Angle
39.
diagram,
same
1 +
7
(c) Angle
angle 1.
Ex.
the
40.
angle
Show
41.
Ex.
The
solution
(1)
adjacent complementary
angles
form
of the
bisectors
of two
adjacent supplementary
other.
of two
pairs of
vertical
angles are
lar
perpendicu-
other.
to each
79.
each
bisectors
The
42.
relation
the
angles to
Ex.
of two
of 45".
student
of
a
The
will observe
theorem
statement
consists
of
the
hypothesisand
of the
Prop. I
from
that the
of four distinct
complete
steps :
ment
theorem, including a general state-
conclusion.
to
possible),
satisfythe conditions set forth in the hypothesis.
of the hypothesis
(3) The application,i.e. the restatement
and conclusion
figurejustdrawn,
as
appliedto the particular
To prove."
and
Given
under the headings of
(4) The proof, or demonstration, the argument by which
(2)
The
drawing
of
a
figure (as general as
"
"
the
truth
or
falsityof
for each
reason
80.
The
1.
and
statement
established,with
is
the
step in the argument.
student
step in
every
theorem.
The
a
"
the
should
be
which
argument
only reasons
Something assumed,
careful to quote the
admissible
i.e.
he
are
makes
of two
in
for
reason
proving
kinds
a
:
definitions,axioms, postulates,
hypothesis.
2. Something previouslyproved,i.e. theorems, corollaries,
and problems.
the
81.
The
necessity
for
proof.
21
I
BOOK
Some
theorems
by merely looking at the figure,and the student
The eye, however,
think a proof unnecessary.
reasoning enables
error, and
detect
The
danger
to be
us
trusting the
of
eye
evident
seem
sure
will doubtless
cannot
always
of
sions.
conclu-
our
is illustrated
in the
followingexercises.*
the
diagrams given below, tell which
by careful
6, and verifyyour answer
In
43.
Ex.
longer,a
line of each
the
or
3
pair is
measurement.
/^k7
c
E1^
b
Ex.
In
44.
apart ?
the
figuresbelow,
Verify
your
answer
are
the
lines everywhere
by using
a
ruler
or
a
the
tance
dis-
same
slipof paper.
//////////////
wwww
A 7Wt"tL///////
Ex.
the
In
45.
other lines.
Verify
figuresbelow, tell which
your
lines
are
prolongationsof
answers.
A
/
z
7
*
These
diagrams
etry, Proceedings
Mathematics
7
of
Teachers.
are
the
taken
by permission from
Eighth Meeting of the
the
Report
Central
of the
Committee
Association
of
on
Geom"
Science
and
22
PLANE
GEOMETRY
POLYGONS.
82.
finite
a
83.
a
84.
is
composed
The
the
straightlines only.
86.
Def.
Any
the
intersection
the
polygon is
brevity,an
BCD,
angle
of
lines
polygon is
a
two
its perimeter.
consecutive
clockwise
interior
of
vertices
boundary lines included
of the polygon.
sides
by
the
are
boundary
around
of
angle
sides
the
the
and
perimeter
polygon,or,
Fig. 1,
BE
CDE,
A,
interior
of
the
polygon.
Def.
If
angles
the
are
are
EAB
87.
an
remaining portion.
plane figurecomposed of
the plane bounded
by it.
of the
segments
called
In
polygon.
ABC,
of the
of the sides of
sum
the
if it separates
closed
plane closed figurewhose
angle formed
right in passing
the
on
and
a
adjacentvertices
The
for
is
a
portionof
of
Def.
a
figure is
closed
be
to
plane from
polygon
85.
of
the
is said
A
polygon, and
found
plane
a
the finite
points of
between
A
plane
line and
Def.
on
portionof
A
Def.
closed
line
A
Def.
TRIANGLES
any
side
polygon is prolonged
through a vertex,the angle
formed
tion
by the prolongaof
a
and
adjacentside
the
is called
polygon.
In Fig.1, A 1, 2, 3, 4, and
88.
Def.
polygon is
89.
Def.
A
line
called
A
a
5
are
joiningany
diagonal
; as
polygon which
an
exterior
exterior
angle
of
angles.
two
non-adjacentvertices
AC,
Fig. 1.
has
the
all of its sides
of
a
equal is
an
anglesequal is
an
equilateral polygon.
90.
Def.
equiangular
A
polygon which
polygon.
has all of its
91.
is
92.
regular
a
93.
of five
one
angular
equi-
a
sides,a
triangle ;
pentagon
;
one
one
on.
so
CLASSIFIED
RESPECT
WITH
trianglehaving
A
Def.
; and
hexagon
TRIANGLES
equilateraland
sides is called
three
sides,a quadrilateral ;
sides,a
is both
polygon.
polygon of
A
Def.
of four
of six
polygon which
A
Def.
23
I
BOOK
sides
two
no
TO
SIDES
equal is
scalene
a
triangle.
94.
trianglehaving two sides equal is an isosceles
equal sides are spoken of as the sides * of the
the equal sides is the vertex
angle between
side oppositethe vertex
angle is called the base.
A
Def.
The
triangle.
triangle. The
angle,and the
95.
its three
triangle having
A
Def.
sides
equal
is
an
equilateral triangle.
Scalene
Isosceles
TRIANGLES
96.
Def.
The
angle.
The
CLASSIFIED
A
WITH
right triangle is
side
other
opposite the
sides
two
are
Equilateral
RESPECT
TO
ANGLES
trianglewhich
rightangle is called
a
spoken
of
as
the
has
a
the
sides
right
nuse.
hypote-
t of the
triangle.
97.
obtuse
98.
Def.
*
The
obtuse
triangle is
a
trianglewhich
has
an
angle.
Def.
angles are
will be
An
t Sometimes
acute
triangle is
a
trianglein
which
all the
acute.
equal
used
An
sides
are
sometimes
occasionallyin
the
of thef triangle. This
sides
term
of
the
a
called
the
arms
of the
isosceles
triangle. This
term
exercises.
right triangle including the
will be found
in the exercises.
right angle
are
called
the
arms
24
PLANE
Ex.
46.
Draw
and
acute
Ex.
with
47.
99.
is
be
of its
The
side
angles
side
considered
to stand
a
with
triangle: (a) with
right angle ; (")
one
which
its
base
upon
of its
one
any
The
from
trianglehas
It will be
can
polygon may
a
the
side may
be
angle,
vertex
of the
triangle.
as
in
Thus
be drawn.
can
to
a
triangleEFG,
if FG
is taken as base, EH is the
altitude; if GE is taken as base,
.Fif will be the altitude;if EF
is taken as base,the third altitude
supposed
to
triangleis the perpendicular
In general any side of a
opposite vertex.
considered
be
Thus
of
altitude
the
triangle may
its base.
obtuse.
its base.
as
Def.
its base
is
zontal
hori-
arms
angle
sides,any
angle oppositethe base of a triangleis
the vertex of the angle is called the vertex
100.
its
one
since
The
and
right angle.
of
with
angles
all its
one
polygon
a
however,
;
with
(a)
:
horizontal; (b)
upon
usuallycalled
supposed
triangle freehand
isosceles
an
one
Def.
stand
its shortest
Draw
and
scalene
a
GEOMETRY
g
^\
^
f/'
/^
^--
gr
every
FlG#
L
three altitudes.
proved
from
be drawn
later that
a
point to
and only one,
perpendicular,
one
line.
a
triangleare often designated by the
small letters corresponding to the capitalsat the opposite
vertices ; as, sides e, /, and g, Eig. 1.
The
101.
Ex.
Do
they
Ex.
Do
48.
they
Ex.
102.
Draw
an
to
meet
Draw
an
seem
49.
meet
50.
of
sides
in
a
Def.
vertices of the
The
triangle;
acute
in
a
point ?
obtuse
the
three
medians
triangleto
is this
draw
is this
Where
altitudes of
of
the
point located
its three
point
a
altitudes
located
freehand.
?
freehand.
?
righttrianglemeet
triangleare
mid-pointsof
a
altitudes
its three
draw
Where
triangle;
point ?
do
Where
a
?
the lines from
the
the
oppositesides.
26
PLANE
GEOMETRY
Proposition
II.
Theorem
triangles are
equal if a side and
to a
adjacent angles of one are equal respectively
the two adjacent angles of the other.
Two
105.
A
Given
To
and
ABC
A
prove
A
ABC=
Z
AC=DF"
DEF,
Z
A
so
A
that
its
C
ABC
A
A
upon
be
to
another
change
of size
D,
F.
Then
linear
will
Also
or
CB
will
its
on
become
with
FE,
4.
.*.
point B
point E.
5.
.'.
A
ABC
its
on
=
must
A
Zf.
A
=
one
tion
posi-
without
or
shape.
14.
Z D,
by hyp.
B
on
coland
fall somewhere
or
=
tion.
prolonga-
will
linear
FE,
and
DE,
fall somewhere
DE,
3.
with
Z
col-
will become
AB
from
moved
"64,
2.
C
Any geometric figuremay
DEF
shall fall upon
equal DF,
upon
side and
Reasons
upon
A C
Z
two
DEF.
Argument
L. Place
D, and
the
3.
Z'C=ZF,
by hyp.
B
on
tion.
prolongafall
on
DEF.
Q.E.D.
intersectingstr. lines
determine
a point. " 26.
5. Two
geometric figuresare
be
equal if they can
4.
Two
made
to coincide.
" 18.
In
53.
Ex.
figurefor Prop. II, assume
the
angle
and
angle Z",
marking the
27
I
BOOK
B
E
angle
=
colored
lines with
;
AB
repeat
the
as
soon
crayon
DE,
=
angle
A
=p
proof by superposition,
as
their
positions are
determined.
Ex.
"
II
polygon
upon
part of
in
Place
54.
that
so
its
I shall fall upon
II.
Discuss
I
polygon
the
some
equal
resulting
positionsof the remaining parts
figure.
of the
Ex.
of
one
other,
have
quadrilaterals
If two
55.
equal
and
respectivelyto three
arranged
in
the
three sides and
sides and
order,
same
the
are
the included
angles
included
the
angles of the
quadrilateralsequal ?
Prove.
106.
of
superpositionshould be used for proving
fundamental
sary
propositionsonly. In proving other propositionsit is necesshow
to
merely that certain conditions are present and to quote
state
conclusions
theorems, previously proved, which
regarding such
The
Note.
method
conditions.
Ex.
the
If at any
56.
bisector
is drawn
thus formed
Ex.
of two
line
are
will be
If
57.
laid off
at the
equal triangleswill
Ex.
to
will
off
cut
bisects
the
be
on
ends
it
are
of the
one
of
measured
these
angle a perpendicular to
an
sides of the
from
angle, the
the
lines,and
segments,
if
two
triangles
point of intersection
perpendicularsto this
two
"
formed.
If at the ends
58.
the
meeting
equal.
equal segments,
lines,are
drawn
the bisector of
point in
straightline perpendiculars
these
drawn,
perpendiculars
equal segments
given line and
of
a
upon
is
any
not
line which
perpendicular
to it.
Fig.
Ex.
59.
angles are
Ex.
60.
Fig.
1.
angles of
equal (Fig.1).
If two
If two
to the bisector
eqjjal
a
triangleare
equal, the
trianglesare equal, the bisector of
of the corresponding angle of the
bisectors
any
other
of these
angle of one
(Fig.2),
is
28
PLANE
GEOMETRY
Proposition
107.
III.
Theorem
Two
cluded
triangles are equal if two sides and the inangle of one
are
equal respectivelyto two sides
the included
angle of the other.
%nd
A
Given
To
and
ABC
A
prove
ABC
=
MNO,
A
AB
=
AG
MN,
=
Place
so
A
ABC
that
AC
its
A
Z
=
M.
Reasons
A MNO
upon
shall fall upon
equal MO,
A
upon
1.
Any geometric figuremay
be moved
from
to
another
change
of size
M,
0.
C upon
Z
MNO.
Argument
1.
MO, and
tion
posi-
one
without
or
shape.
" 54, 14.
Then
2.
linear with
Point
3.
4.
*
.
.
B
col-
will become
AB
ZA
3.
AB
4.
Only
=
will fall on
pointN.
NO.
=
be
5.
ABC=A
A
ZM, by hyp.
MN.
will coincide with
B C
2.
5.
MNO.
MN,
by hyp.
one
str.
drawn
line
between
can
two
points. " 25.
Two
geometric figuresare
equal if they can be made
Q.E.D.
to coincide.
" 18.
right triangles are equal if the two sides
including the right angle of one are equal respectively
to the two
sides including the right angle of the other.
108.
Ex.
Cor.
61.
Two
Prove
Prop. HI
by placing AB
upon
MN,
109.
which,
one
In
Def.
lines,and
Hence
29
I
BOOK
the points,
lines,and angles in
equal figures,
superposed,coincide respectivelywith points,
parts.
angles in the other, are called homologous
when
:
bisect each
straightlines
equal in pairs.
If two
62.
Ex.
equal.
Homologous parts of equal figures are
110.
extremities
are
other,
the
lines
joiningtheir
anglesequal,try to find two triangles,
of the angles. If the trianglescan
of the lines or one
each containingone
lines or two
be proved equal, and the two
angles are homologous parts of
the triangles,
then the lines or angles are
equal. The parts given equal
be more
symbol,
by marking them with the same
easilyremembered
may
To
Hint.
with
or
Ex.
what
prove
colored
the sides of
an
the bisector
of
Ex.
to
which
Ex.
the
they
of
arms
segments
to
an
the
medians
drawn,
are
If
66.
are
67.
laid off
the
oppositeends
Extend
on
you
make
of
perpendicular to
are
Prove
?
each
other,
its correctness.
triangleare perpendicular to
triangleis equilateral.
a
equal segments measured
isosceles triangle,the
Fig.
Ex.
can
62
from
laid off on
the vertex
are
equal segments measured
angle, and if their extremities are joined to any point in
the angle, two
equal triangleswill be formed.
If two
65.
lines in Ex.
statement
If
64.
the
case
additional
Ex.
two
or
crayon.
In
63.
lines
two
the
of the base
from
lines
will
the vertex
Fie.
arms
66
to
the
case
in
prolonged through
laid off
joining the ends
be equal (Fig.1).
1.
Ex.
are
which
the
vertex
the sides
on
of these
2.
the
equal segments
(Fig.2).
30
PLANE
GEOMETRY
Proposition
111.
The
base
IV.
angles of
Theorem
isosceles triangle
an
are
equal.
B
isosceles A
Given
To
112.
an
Z
prove
Cor.
I.
isosceles
bisects
113.
Ex.
eaual.
A
=
Z
The
with
ABC,
AB
and
BC
its
equal sides.
C.
bisector of the angle at
triangle
is
perpendicular
to
the vertex
the
base
of
and
it.
Cor.
68.
The
n.
An
bisectors
equilateral triangle
of
the
base
angles of
an
is
also
isosceles
angular.
equi-
triangleare
114.
in
of asses, since it has
bridge
or
norum,
This
geometry.
here
and
the
are
due
is
69
proved
known
as
the
asi-
pons
difficult to many
beginners
proposition
suggested
proof
to
Euclid,
who
mathematician
Exercise
Note.
Historical
31
I
BOOK
great
a
the
wrote
on
systematic text-book
geometry. In this work, known
first
Elements, the
Euclid's
as
here
given
in Book
Of
"
I.
the life of Euclid
was
taught
in Alexandria
b.c."
To
him
"
There
saying,
to
geometry.
lived
and
about
300
is attributed
is
' '
of the culture
he
and
modest
who
Greek
a
there is
that
except
gentle and
was
fifth proposition
is the
littleknown
but
cise
exer-
the
royal road
no
Euclid
His
appreciation
value
of geometry
in
is shown
lad who
(which is probably authentic). "A
I
do
asked, What
gain by learning all this
had
*"
slave
and
said,
profitout
of what
'
Give
this
boy
he learns.'
some
story related
a
just begun geometry
'
stuff ?
coppers,
by Stobaeus
since
Euclid
he
called
his
must
make
a
are
laid off
"
By using the accompanying diagram prove
that the base angles of an isosceles triangleare equal.
Ex.
69.
Hint.
A ABE
Prove
A ACE
vertex
laid off
are
the lines drawn
of the
bisector
(6)
the
laid off
the
on
from
=
ABAC.
arms
the ends
of the vertex
Extend
BBC.
Then
of
an
of the segments
angle
to the
the
arms
of
an
isosceles
from
isosceles
will be
the
triangle,
to the foot
equal.
the
(a) If equal segments measured
71.
on
prove
measured
in which
case
(a)
the
vertex.
prolonged through
arms
Ex.
A
(a) If equal segments
70.
Ex.
=
equal segments
from
triangle,the
the ends
lines drawn
of the segments to the foot of the bisector of the vertex
of the base
from
are
are
the ends
angle will
(6) Extend
(a) to the case in which the equal segments
the arms
prolonged through the ends of the base.
on
be
equal.
laid off
on
32
PLANE
Ex.
72.
are
of the
the
on
equal segments
base
triangleto the
Extend
(6)
the
If
(a)
laid off
of
Ex.
are
Extend
(6)
segments
Ex.
115.
CD.
=
in
base
ABC
Prove
from
the
ends
lines drawn
of the
will be
base
2.
of the
base
from
the
equal.
prolonged
(Fig.2).
equilateral,and
is
triangle EFD
theorems
lateral.
equi-
by
which
Means
trianglesequal
prove
in measuring
applied practically
distances
on
the
surface
the earth.
Thus, if it is desired to find the distance
two
places,A and B, which are separated by a pond
obstruction,
placea stake at some point
accessible
Measure
to
both
and
A
the distances
FA
keeping in line with F
CF
equal to FB, and,
and
A,
measure
AB
on
is
measure
CE, and
FE
B,
and
and
in
equalto
FB
B,
Can
a
;
FA.
of
between
or
other
then,
measure
F
Lastly
from
A
this method
hill and
are
F.
as
line with
the distance
equal to CE.
oppositesides of
on
equal
the
arms
of Distances
The
laid off
are
triangle,the
which
the
on
Measurement
Triangles.
isoscles
an
case
off
vertex
equal.
equal segments
measured
the opposite ends
the
to
Triangle
74.
BF
=
of
of the
ends
the
joiningthe
base
.
equal segments
to
laid
lines
of the
Fig.
arms
(a)
are
the
through
of
the
on
will be
the ends
1.
of the segments
ends
AE
If
(a)
73.
laid off
in which
case
from
the
triangle,
of the segments
to the
Fig.
measured
isosceles
an
ends
(a)
prolonged (Fig. 1)
base
GEOMETRY
each
to B
is thus
be used
obtained,since
when
is invisible from
A
and
B
are
the other ?
34
PLANE
GEOMETRY
Proposition
116.
Two
triangles
are
V.
Theorem
equal if
the three
sides
of
to the three sides of the other.
equal respectively
are
A
Given
To
and
ABC
A
prove
ABC
=
BC
RST, AB=RS,
A
ST, and
=
A
RST
longestside
its
upon
A,
upon
so
that
S
TR.
=
Reasons
the
so
that
RT
shall fall
equal
T
"
RST.
Argument
Place
CA
one
upon
AC,
1.
Any geometric figure maybe
moved
to
R
from
another
change of
" 54, 14.
C, and
shall fall opposite
tion
posi-
one
without
size
or
shape.
B.
2.
Draw
BS.
2.
str. line
A
from
any
AB
=
The
A
Same
If
drawn
point
" 54, 15.
RS, by hyp.
any
other.
base A
are
BC"ST,
be
may
one
of
isosceles
an
equal. "
by hyp.
reason
as
equals are
equals, the
to
111.
4.
added
sums
equal. " 54?2.
to
are
BOOK
35
I
Reasons
Argument
8.
Z
.*.
8.
ABC=/-CSA.
The
whole
all its
9.
A
.'.
ABC
i.e. A
A
=
ABC
=
A
A
9. Two
CSA;
parts.
of
one
the
Ex.
81.
(a)
the shortest
(b)
side
side of
Prove
of
117.
to
one
the
shortest
Prop. V, using two
SUMMARY
is not
Why
Question.
CONDITIONS
OF
side of the
two
Z
included
sides
and
of
the
Z
" 107.
trianglesand applying
other.
righttrianglesand applying the
shortest side of the
to the
one
obtuse
Prop. V, using two
Prove
if two
equal respectively
are
other.
Q.E.D.
" 54, 11.
the included
to
of
sum
equal
are
sides and
RST.
the
=
shortest
other.
Prop.
FOR
V
proved by superposition?
EQUALITY
a
OF
side and
TRIANGLES
the two
adjacent
angles
trianglesare
Two
118.
equal if
two
sides and
the included
angle
three sides
a
the two
side and
adjacent
angles
of
one
to
equal respectively
are
two
sides and
the included
angle
three sides
of the other.
.
Ex.
the
angle at
Ex.
The
median
vertex
and
82.
In
83.
other
two
prove
equal.
sides
a
is
certain
are
trianglebisects
the
quadrilateraltwo adjacent sides are equal ;
equal. Find a pair of triangleswhich you
the
to
also
the
base of
an
perpendicularto
isosceles
the base.
can
36
PLANE
Ex.
If the
84.
angles also
Ex.
opposite sides
equal.
are
If two
85.
their
GEOMETRY
isosceles
vertices bisects each
base.
of
quadrilateralare
a
triangleshave
vertex
angle
and
three
is
base, the line joining
same
perpendicular to
the
mon
-com-
(Two cases.)
Ex.
86.
In what
trianglesare
the
Ex.
87.
In
trianglesare
two
Ex.
88.
If three
triangle,can
different
the
equal,the opposite
what
rods
different
a
order
medians
equal ?
equal
?
put together to form
lengths are
a
by arranging the rods in a
trianglebe formed
the angles opposite the same
rods always be the
Will
?
of different
medians
?
same
Ex.
If two
89.
sides of
sides
another, and
the
is
is
a
such
curve
within
A
Def.
to the
is
all
that
drawn
median
circle
triangleare
one
drawn
equal to the median
trianglesare equal.
119.
of
to
one
equal respectivelyto
two
of these
first
sides in the
in the
corresponding side
plane closed figurewhose
straightlines to it from a
a
second,
boundary
fixed point
equal.
are
boundary of a circle is called
The fixed point within is called the center,
circumference.
to any point on the circumference
a line joiningthe center
The
the
which
curve
forms
the
the
and
is
a
radius.
the definition
It follows from
120.
All radii
of the
circle
same
are
Any portionof
121.
Def.
122.
Assumption
a
Circle
18.
of
circle that
:
equal.
circumference
postulate.
having any point as center, and
any
a
A
is called
circle may
having a
radius
an
be
arc.
structed
con-
equal to
finiteline.
123.
three
The
distinct
(1)
Tlie
The
which
problem of
construction
consists
of
steps :
ruler and
proof
fulfills the
(3)
a
i.e. the process of drawing the required
construction,
figurewith
(2)
of
solution
The
there
a
compasses.
demonstration
that
the
figure constructed
given conditions.
discussion,i.e. a
may
be
no
statement
solution,one
of the
conditions
solution,or
more
under
than
one.
BOOK
line
construct
To
Problem
VI.
Proposition
124.
37
I
equilateraltriangle,with
an
a
given
side.
as
A
line
Given
To
BC.
construct
an
equilateral
triangleon
BC.
I. Construction
1. With
B
as
center
and
BC
as
2. With
C
as
center
and
BC
as
3.
with
4.
Connect
and
B
A
point A,
at which
the
is the
ABC
BC
=
.'. AB
Proof
3Hid CA
Reasons
"
All radii of the
BC.
=
2.
BC=CA.
equal.
.'.A
ABC
is
equilateral.
This
construction
solution.
is
(See" 116.)
A A
circle
" 120.
same
each
" 54, 1.
having
its three sides
" 95.
equalis equilateral.
Q.E.D.
III.
same
Things equal to the
thing are equal to
other.
3.
intersect,
requiredtriangle.
are
2.
circle BMA.
circumferences
Argument
AB
CAB.
C.
II.
1.
circle
radius,construct
radius,construct
Discussion
and
always possible,
there
is
onlyone
38
PLANE
GEOMETRY
Proposition
With
125.
VII.
Problem
given vertex and
angle equal to a given angle.
an
and
A, side AB, and
vertex
Given
To
a
construct
AB
Z
an
2.
With
D
With
A
an
4.
Draw
5.
Z
and
CDE
with
center,and
as
having
IH, cuttingAB
sides of Z
at F
D
same
I as
center,and with
arc
the
intersecting
a
arc
radius
=
Z
CDE, and
is the Z
DG
4.
FG
"
=
radius,describe
the
str. line
FG,
required.
Proof
Reasons
FG
and
1. A
IK.
str. line may
any
3.
G, respectively.
at K.
IH
from
In A
scribe
radius,de-
and
equal to
Argument
2.
vertex
AK.
BAK
Draw
as
at /.
II.
1.
A
convenient
any
center,and with the
as
arc
With
describe
CDE.
Construction
intersectingthe
arc
indefinite
3.
construct
side.
as
an
Z
equal to
I.
1.
Z
given side, to
a
any
other.
2.
By
cons.
AK.
3.
By
cons.
IK.
4.
By
cons.
FDG
and IAK,
DF"AI.
be
one
drawn
point to
" 54, 15.
39
I
BOOK
Eeasons
Argument
5.
.*
.
A
FDG
A
=
5.
IAK.
A
Two
three
if the
equal
are
sides
of
one
are
to
equal respectively
the
sides of the other.
three
"116.
6.
."
ZBAK
.
6.
Z.CDE.
=
Homol.
figuresare
Q.E.D.
III.
This
is
construction
of
parts
equal
equal. "
110.
Discussion
always possible,and
is
there
only one
solution.
triangle,given two sides and the included
Ex.
90.
Construct
Ex.
91.
Construct
an
Ex.
92.
Construct
a
93.
Construct
an
a
triangle,given
isosceles
triangle,given
side
a
the
and
angle and
vertex
the
angle.
adjacent
two
angles.
Ex.
the
How
94.
determine
it ?
parts determine
many
an
arm
and
Do
three
of
one
95.
Construct
an
Ex.
96.
Construct-
a
from
The
Def.
the vertex
97.
In
Ex.
98.
In
and
bisectors,
Ex.
99.
medians,
In
the
scalene
of the
what
the three bisectors
isosceles
bisector
by the opposite
triangle.
Ex.
a
triangle?
angles
Explain.
Ex.
126.
the
given
triangle,
equal angles.
Ex.
two
isosceles
side
given the
triangle,
triangle^given the
of
an
of
a
a
triangles are
triangles are
only two, equal ?
triangles are
bisectors,and the
an
arm.
sides.
triangle is the
the
what
altitudes identical ?
of
three
angle bisectingthe angle
equal ?
what
angle
and
base
and
line
limited
40
PLANE
GEOMETRY
Proposition
127.
the bisector
To construct
the bisector
construct
I.
1. With
describe
2.
arcs
B
an
With
arc
D
Draw
4.
BF
of Z
of a given angle.
with
BA
intersecting
and
E
as
ABC.
Construction
center, and
as
at
intersecting
3.
Problem
Z.ABC.
Given
To
VIII.
at E
centers, and
F.
BF.
is the bisector
of Z
ABC.
II.
Proof
convenient
any
and
with
BC
radius;
at D.
equal radii,describe
42
PLANE
GEOMETRY
LOCI
In
129.
geometric problems it is necessary to locate
all pointswhich
satisfycertain prescribed conditions,or to
determine
the path traced by a point which
moves
according
to certain ijxed laws.
Thus, the points in a plane two inches
from a given point are
in the circumference
of a circle whose
is the given point and whose
center
radius is two inches.
Again, let it be required to find all points in a plane two
inches
from
fixed point and three inches from
one
another.
All
cumferen
points two inches from the fixed point P are in the cirof the circle LMS, having P for center and having
radius
a
equal to two
many
inches.
All
inches
from
point
Q
pointsthree
the
in
are
fixed
the
cumference
cir-
of the circle
having Q for center
and
having a radius
equal to three inches.
LET,
If
the
circles
two
wholly outside
the two
are
of each
prescribedconditions
but do not
will be
other,there
; if the two
there will be
intersect,
there
intersect,
one
will be two
(" 324) that there cannot be more
satisfyboth of the given conditions.
other
A
circumferences
touch,
than
two
pointswhich
figureis the locus of all points which satisfy
or
more
given conditions,if all pointsin the figuresatisfy
given conditions and if these conditions are satisfied by no
130.
the
pointssatisfying
ferences
point; if the two circumpoints. It will be proved
later
one
no
A
Def.
points.
locus, then, is
an
assemblage of points which
obey
one
or
definite laws.
more
It is often
them
as
the
is controlled
convenient
path
traced
by
certain
locate
to
by
a
these
pointsby thinking of
moving point the
fixed laws.
motion
oi which
plane geometry a
more
pointsor of one or more
pointsand lines.
132.
Questions.*
a
the
between
jet four
from
Given
109.
Ex.
which
in AB
a, from
are
Given
110.
pointsfive
radius
of
locus
feet from
a
gas
the
ceiling
ceiling,five feet
the
wall ?
and
point P.
P; (b) at
all
Find
points
given distance,
a
from
inches
or
:
from
radius
a
of
(a)
center
0 and
all points four
circumference
the
radius six inches.
inches
of the
from
State,
0 ;
(")
measured
circle,
of
on
prolonged.
Given
of the
locus
circle with
a
inches
111.
Ex.
the
(a)
:
locus
proof,the
without
the
three
four
feet from
end
an
equator
is the
midway
23"" from
?
P.
Ex.
all
also
of
of the earth
the
What
?
line AB
unlimited
an
surface
room
four
six feet from
side wall, and
a
wall ?
side
a
combination
all
from
?
equator
ceilingof this
the
from
the
of
the
poles? 23|"
south
from
00"
from
feet
five feet
and
and
north
pole ?
north
the
or
one
points in space two
is the locus of all points in space
two
in space such
is the locus of all points*
a given plane shall be equal to a
given
locus
given point ? What
What
inches from a given plane ?
them
to
that perpendicularsfrom
is the locus of all points on
What
line ?
from
inches
of
composed
of any
lines,or
is the
What
"
be
locus may
In
131.
43
I
BOOK
the base and
of the
vertex
adjacent angle of
one
angle oppositethe
base
a
?
what
is
triangle,
(Statewithout
proof.)
Given
112.
Ex.
locus of the vertex
is the locus of the
is the
locus of the
median
to
the base
133.
*
In
this
v
*
the
base, what
In
Question.
base
two
sides of
angle oppositethe
triangleand
a
median
is the
other
?
of
which
base
to the
from
triangle,one
other
locus of the vertex
of the
a
triangle,what
?
the median
is remote
a
the
base, what
base
?
side, and the
angle opposite
which
of
the
exercises
above
was
a
triangle
?
order
to
article
proofs of
no
the
of
the
base
what is the
triangle,
(Statewithout proof.)
a
?
determined
in
Given
115.
Ex.
of the
side of
other
one
and
of the
the base
end
and
angle oppositethe
the base
vertex
Given
114.
Ex.
of the
Given
113.
Ex.
the base
to
answers
the
develop
introduce
to
imagination
of the
student
loci
questions involving
questions are demanded.
these
in
the
space.
authors
deem
It should
it advisable
be
noted
that
44
PLANE
GEOMETRY
Proposition
Every
134.
line
is
point
in
IX.
perpendicular
the ends of that
C
A
line AB, its "
To
PA
prove
AC
2.
as
bisector CD, and
Z
4.
.*.A
line.
P
pointin
any
CPB,
1.
By hyp.
2.
By
CB.
iden.
PC A
=
Z BCP.
3.
All rt. A
APC
=
A
4.
Two
CPB.
A
equal. "
are
equal
are
sides and
of
Z
the
are
one
to
and
.\PA
are
Q. E. D.
Ex.
116.
Four
C 20 miles north
far from
Ex.
two
C
117.
as
of
villagesare
A,
it is from
In
a
given points. A
and
D
so
20
located that B
miles
south
equal
spectively
re-
sides
two
Z
of
" 107.
equal. "
is 25
of A.
miles east
Prove
that
B
110.
of
is
A,
as
D.
given circumference,find
and
included
ures
parts of equal fig-
5. Homol.
PB.
=
64.
if two
included
the
the other.
5.
CD.
REASON8
PC=PC.
3.
a
PB.
and
APC
of
B
Argument
1. In A
bisector
the
equidistant from
Given
Theorem
B.
the
points equidistantfrom
BOOK
135.
of the first is the
conclusion
hypothesis of
The
of
false.
truth
a
is true, but
bipeds
136.
is not
the
the
of the
theorem
One
Def.
the
All
"
converse,
true ;
thus, " All
The
of
opposite
lives in the
York,
137.
is true
is not
New
York,
opposite, If a
equal
when
the
hypothesis
tion
contradic-
does
live in the
not
State,"is
York
propositionis true,
a
"
thus, If a
York
State,"
true ;
he lives in New
man
of
converse
too, if the opposite of
; so,
first is the
always
live in Neiv
not
If the
Note.
truth
a
"
he does
of the
of the
is
of the second.
cityof
is true, but the
"All
is the opposite of another
men
men,"
bipeds are
first is the contradiction
of the conclusion
the
of the second.
always
second, and the conclusion
when
second, and the
right angles are equal" is true, but
rightangles" is false.
hypothesisof
New
hypothesisof
"All
angles are
man
of another
converse
the first is the conclusion
converse
"
are
is the
theorem
One
Def.
45
I
false.
the
proposition is true, the
a
cityof
opposite also
also
converse
is true.
This
be evident
may
type forms
to the
Then
If
must
138.
^lis.B,
C is D.
Then
A
(3) must
of the
lowing
fol-
(8) Opposite
CisD,
If
A
A
Then
C is not
is B.
be
is not
B,
D.
Again, if (3) is true, then
true.
is that
oppositeas
well
State
118.
straightline
119.
;
its
be
sufficient test of the
and
necessary
definition may
Ex.
consideration
(2)
true.
definition
Ex.
a
(2) Converse
If
(2) is true, then
be
after
:
(1) Direct
If
student
quoted
for
as
the
shall
converse
a
as
the
also
for
reason
direct statement
of
converse
completenessof
be
a
in
converse
Hence
or
for
a
an
argument.
an
definition
the
true.
for
equal figures;
plane surface.
State
straightline,the
sum
the
converse
of the two
of
If
another
straightline meets
adjacent angles is two rightangles.
Ex.
120.
State the
converse
and
Ex.
121.
State the
converse
of
:
one
opposite of Prop.
Prop.
I.
a
Is it true?
IX.
46
PLANE
GEOMETRY
Proposition
X.
Theorem
(Converseof Prop. IX)
Every point equidistant from the ends of
the perpendicular bisector of that line.
139.
lies in
line RS, and
Given
To
such, that
QR
Q lies in the _L bisector
that
prove
point Q
Reasons
bisect Z
QT
QS,
=
of RS.
Argument
1. Let
line
a
Every
RQS.
Z
but
has
sector.
bi-
one
" 53.
2.
QR
3.
.'.A
By hyp.
QS.
=
RQS
A A
is isosceles.
having
is
4.
is the
QT
.'.
_L bisector
of
The
an
bisector
A
is "
of
to
bisects it.
5.
.'.
Q
lies in the _L bisector
of
RS.
140.
Cor.
of
a
5.
of
equal
" 94.
isosceles A.
the vertex
RS.
sides
two
Z
the
isosceles
an
the
at
base
and
" 112.
By proof.
Q.E.D.
I.
line
Every point
is not
not
in
the
equidistant from
perpendicular
the
ends
sector
bi-
of the
line.
Hint.
Use
" 137, or
is false.
contradict the
conclusion
and
tell
why
the
tradiction
con-
141.
the ends
that
of
given line is the
a
pointsequidistantfrom
perpendicularbisector of
line.
(For
locus theorems,
proofsof
model
142.
the
" 139 and "
Hint.
Use
143.
In order
298.)
perpendicularbisector of the
ends
line.
25.
that
to prove
to show
sufficient
and
necessary
pp. 297 and
see
pointseach equidistantfrom the
Two
Cor. in.
line determine
of a
all
of
locus
The
Cor. n.
47
I
BOOK
problem is
locus
a
solved it is
things:
two
(1) That every point in the proposed locus satisfies the
conditions.
prescribed
(2) That every point outside of the proposedlocus does not
conditions.
the prescribed
satisfy
convenient
be more
it may frequently
Instead of proving(2),
to prove
:
(2')That
every
lies in the
144.
a
In exercises in which
Note.
locus,"
ditions
pointwhich satisfies the prescribedconproposedlocus.
understood
be
it must
that
given a proof with regard to it as
and
based upon a direct proposition
he
the
student
has
not
is asked
found
outlined above.
has
its
opposite;
locus until he
a
The
or, upon
"Find
to
proof must
a
be
direct proposition
and its converse.
and
pointsA
Ex.
two
given unlimited
a
at the
Ex.
125.
126.
not
two
given
(6)
a
distance of
a
Given
a
from
pointequidistant
a
of circle
m
circle 0 with
from
O,
a
point
and
which
a
P.
are
Find
also
givenpointP,
givenpoint Q.
radius
all
(a)
P.
of d from
distance
a
also
0,
center
distance of d from
all pointsat
Find
find
AB,
this line.
on
circle with
a
P ;
time at
same
points equidistantfrom
line
lie in the circumference
from
inches
Ex.
Given
124.
pointswhich
of all
B.
In
123.
locus
given pointsG and D
Ex.
two
the
Find
122.
Ex.
r.
Find
a
and
points
the locus of the mid-
of the radii of the circle.
Ex.
127.
proof,the
Ex.
Given
locus of
128.
passes
a
having the
from
pointequidistant
two
circles
perpendicularbisector
through the vertex,
The
State,without
center.
same
their circumferences.
of the
base
-
of
an
angle
isoscelestri-
48
PLANE
GEOMETRY
Proposition
To construct
145.
the
XI.
Problem
perpendicular bisector of
a
given
straight line.
'H
V
B
A
I
I
\
IF
x
Fig.
line AB
Given
To
2.
(Fig.1).
the
construct
Fig.
1.
bisector
perpendicular
of AB.
I. Construction
1. With
than
A
half AB, describe
2.
With
B
with
center, and
as
the
arc
center, and
as
a
radius
convenient
greater
EF.
with
the
radius,describe
same
the
HK.
arc
Let
3.
and
c
D
be
the
points of
and
D.
intersection of these two
arcs.
4.
Connect
5.
CD
II.
points C
is the _L bisector
The
proof and
of AB.
discussion
left
are
as
exercise for the
an
student.
Hint.
146.
Apply " 142.
Note.
The
construction
positionof the given line makes
147.
ghall
Question.
equal DA
it more
Is it necessary
(Fig. 1) ?
Give
given in Fig. 2 may
the
be used
when
the
convenient.
that
CA
shall
equal CB?
equations that must
hold.
that
CA
50
GEOMETRY
PLANE
Proposition
From
149.
XIII.
point outside
a
Problem
line
a
to construct
a
pendicula
per-
to the line.
*"B
Xq
line
Given
To
and
AB
construct
_L from
a
outside of
pointP
AB.
to AB.
P
I. Construction
With.
1.
describe
P
an
arc
With
2.
describe
c
Draw
4.
PQ
cuttingAB
and
D
is
a
_L from
discussion
will be
150.
that
138.
Ex.
they
do
drawn
to
140.
they
Ex.
Must
with
any
convenient
radius,
seem
141.
to
one
as
an
P
and
exercise
for the
student.
The
154.
Q be
on
opposite sides
of AB?
Is it
PC=QG?
? where
meet
three
altitudes of
an
the three
altitudes of
a
the
acute
triangle. Do
they
?
Construct
139.
seem
Ex.
D.
length,
Q.
given in "
Construct
meet
to
and
sufficient
to AB.
P
is left
Question.
necessary
seem
of
,
proof
Ex.
centers, and
as
radius
PQ.
The
II.
a
points C
at
at
intersecting
arcs
3.
with
center, and
as
righttriangle. Where
do
?
Construct
to meet
three
altitudes of
an
obtuse
triangle. Where
?
Construct
of them.
the
a
triangleABC,
Abbreviate
thus
sides
and
givena, br and
m^.
given
:
two
the
median
51
I
BOOK
I
151.
of
Analysis
difficult
problems of
more
152.
*
Note.
In
such
problem solved
supposed
the student
sketch
to
will
matter
Thus
Problem.
:
isosceles
an
altitude
the
complete
to
it be
one
be
readily
prolong BD
one
and
of the
given sides.
143.
Construct
the
given
a,
Ex.
to
Ex.
which
Ex.
median
Ex.
case
upon
altitude to the
and
B,
144.
desired
the
parts
of the
given
convenient)
relation
struction.
con-
figurethat you
can
struct
con-
and
with
the
colored
solution
the
the
complete
to
Construct
Ex.
the
C, making BC
to
the
easy
to
arm
or
that
this
142.
Ex.
an
(heavy
seen
that
seen
274.
triangle.
usually an
of the
tion
construc-
righttriangle,given the hypotenuse
The
now
right triangleABB
may
arm.
is
studying the figure with
By
crayon), it will be
problem depends in
a
if
part of the
required
triangle,given
given parts marked
of
it is
a
and
required figure.
Let
it.
upon
be
generally
this part is constructed
in " 152
141, it is well first to imagine
crayon,
given. By studying
figure,try to find some
This
construct.
After
colored
(with
times
some-
the process of
of reasoningis called
figure to represent
a
more
to discover
Ex.
as
whole
to the
parts
problems
the
In
reasoning is
carefully the
be
to
and
mark
Then
to
of
course
requiredfigure. This course
analysis of the problem. It is illustrated
fullytreated in the exercises following "
the
the
a
the
drawing
the
construction
to enable
necessary
of construction.
problem
a
=
AB, and
Abbreviate
to
constructed, and
opposite
the
:
it is
given
given
and
b, and
a,
to
C.
sides and
two
it will
only necessary
A
connect
given
thus
triangleABC,
a
side
be
construction
triangleABC,
a
and
altitude
an
hi,.
side,an adjacent angle,
thus :
given angle. Abbreviate
a
ft*.
Construct
an
isosceles
triangle,given
Construct
an
isosceles
triangle,
given
arm
an
and
the
dian
me-
it.
145.
the
median
to
it makes
146.
Construct
to
side
one
147.
its bisector.
a
makes
Construct
a
with
an
and
arm
the
angle
it.
triangle,given two
:
(a) with that side
triangle,given
a
sides
;
(")
side,an
and
with
the
the
angle which
other
a
side.
adjacent angle, and
52
PLANE
GEOMETRY
t
Proposition
If
153.
side
one
is
angle formed
angles.
of
a
XIV.
Theorem
triangle
than
greater
is
prolonged, the
either
of the
B
A
Given
To
with
ABC
Z
prove
DCB
Z
prolongedto
ABCov
Z
Let
mid-pointof
AE, and
draw
BCj
it to F,
=
2.
In A
and
ABE
BE
3.
4.
=
AE
Z
making
Draw
AE.
BE
A
DCB.
CAB.
Reasons
be the
E
exterior Z
making
D,
Argument
1.
terior
in-
remote
F
AC
"
exterior
1.
A
long
proEF
str. line
be drawn
may
from
any
any
other.
one
point to
" 54, 15.
CF.
EFCy
2.
By
EF.
=
Z
point
mid-
of BC.
EC.
=
is the
E
cons.
CEF.
3.
By
cons.
4.
If
two
lines
str.
sect,
inter-
the vertical
equal. "
5.
.'. A
ABE
=
A
EFC.
5.
Two
A
sides
Z of
equal
and
the
the
=
Zfce.
6.
Homol.
included
spectively
re-
sides
two
included
the other.
.-.Zb
if two
equal
are
to
and
are
77.
are
one
A
Z
of
" 107.
ures
parts of equal figare
equal.
" 110.
BOOK
53
I
Reasons
Argument
Z
7.
DCB
Z FCE.
"
The
7.
whole
"
of
any
its
parts. " 54, 12.
8.
Z
.-.
DC"
SubstitutingZ.B for its
equal Z i^C^.
By bisectingline AC, and
by steps similar to 1-8.
8.
5.
is prolonged
if BC
Likewise,
9.
Z
"
9.
G, Z.ACG"
to
Z CAB.
10.
But
11.
.*.
Z
Z
DCB
Z
=
Z)C5
"
ZDCB"
12.
Z
or
154.
there
.4CG.
SubstitutingA DCB
its equal Z.ACG.
By proof.
CAB.
11.
A
ABC
12.
reason
as
4.
for
Q.E.D
From
exists
Same
Z
CAB.
Cor.
10.
only
outside
point
a
line
a
perpendicular
one
to
the line.
If there
Hint.
P,
and
then
PC,
as
are
From
A PC
therefore
A
second
a
154
both
are
this is
But
from
_L to AB
Z. PKA
and
equal.
"" 149 and
155.
follows
exists
rt.
B
"
A
impossible by "
be combined
may
7^ \f
A
in
153-
one
statement
and
only one
as
:
point outside
a
line there exists
a
one
pendicular
per-
to the line,
Ex.
148.
A
Ex.
149.
In the
angle BCA
?
than
Ex.
In
150.
Fig. 1, prove
(3) Angle
(4) Angle
Ex.
greater
EDA
than
greater than
is
necessarilygreater
j"
:
angle
CAE
or
than
angle
CBA
or
;
is greater than
angle 3
angle DAE.
;
Fig. 2, show that angle 6
angle 7 ; also that angle 9
angle A.
In
angle DCB
than
4 is greater than
151.
that
rightangles.
two
figureof Prop. XIV,
than
angle B ?
(1) Angle 1 is greater
angle AEC
;
is
5
(2) Angle
greater
angle BAE
contain
cannot
triangle
is
is
Fig.
2.
54
PLANE
GEOMETRY
Proposition
XV.
Theorem
If two sides of a triangle are unequal, the angle
opposite the greater side is greater than the angle opposite
156.
the less side.
A
Given
To
with
ABC
Z
prove
CAB
BC
"
"
Z
BA.
C.
Argument
1.
On
2.
Draw
AD.
3.
Then
Z.
lay off
BC
Reasons
BD
=
AB.
=Z2.
1.
Circle
2.
Str. line post. I.
" 54, 15.
3.
The
isosceles
base A
A
4.
NowZ2
Z.C.
"
4.
If
post. "" 122, 157.
of
111.
equal. "
are
side of
one
an
is prolonged,
A
a
"
the ext. Z formed
either
"
int. A.
5.
r.Zl
"
Zc.
Z
CAB
"
the
5. Substituting Z
Z 1.
6.
The
remote
" 153.
equal Z
(5. But
of
1
for
its
of
its
2.
whole
"
any
parts. " 54, 12.
7.
/.
ZCAB
"
ZC.
7.
If three
magnitudes
kind
same
that the
and
the
"
q.e.d.
157.
Note.
postulates and
related
so
first "
the
third,then
the third.
the
"
the
first
" 54, 10.
required
definitions in full unless requested to do
the
the
student
not
so
by
ond
sec-
second
be
Hereafter
will
are
of the
to
state
teacher.
BOOK
158.
When
Note.
prolonged
of
of its
the
between
the
other
two
method,
indirect
thus
established.
every
possible contradiction
consider
case, viz.
This
This
of
that
every
XVI
and
161.
as
(a)
argument
of
method
us
will be
Would
Question.
of the
of
a
tively
respec-
the
two
the
propositionis
"
B}
"
B,
=
B.
example,
For
be
to
necessary
in this
admissible
\
\
(3)is established,
of
truth
reasoning is called
of it leads
to
proved by
of
examination
conclusion.
the
a
it be
consists in
ad
by showing
absurdity. Props.
conclusion
an
the indirect
possibleto
reductio
base
a
method.
proof upon
a
diction
contra-
hypothesis?
In the
1, all
use
the
suppositionsshould
As
K.
greater sides.
two
requires an
to establish
of the
indirect
suppositions of
admits, includingthe conclusion.
(")
angle
greater
of the
conclusion
of
contradiction
XVII
BT
then showing
proposition,
a
the
(2) false,
It enables
absurdum.
160.
the
and
between
proof by exclusion,
(i) A
(2) A
(3) A
B.
the base
greater than
included
angle
process
:
By proving(1) and
=
KSR
ST
that A equals B it would
indirectly
that are
the only three suppositions
prove
i.e. A
initial step in developing
quadrilateralare
a
or
The
to be false.
contradiction
with
angle between
contradictingthe conclusion
to
laying
a
sides,the
The
159.
the
as
the
to the ends
point within a trianglelines are drawn
these lines is greater than the angle
angle between
sides of the triangle. (See Ex. 151.)
If from
154.
of
sides, the
two
is greater than
sides
one
other
the
than
Ex.
adjacent sides
If two
angle
Prove
to K.
length,as
any
153.
shorter
serve
triangleEST,
isosceles
the
Given
152.
Ex.
greater will often
given unequal,
are
proof.
a
Ex.
magnitudes
two
the
off of the less upon
55
I
reason
be
As
which
reason
the
1 the
case
he
number
is
give,
considering
of such
ble
possi-
cited.
for the last step in the argument
these suppositionshave
should
the student
method
been
proved
false.
he should
state which
56
PLANE
GEOMETRY
Proposition
XVI.
Theorem
(Converseof Prop. IV)
162.
If
A
To
r
Z.R
with
RST
prove
a
equal,
triangle are
t
=
"
Z.T.
.
Reasons
Argument
1.
r
2.
First suppose
tj r"t,
"
or
r
=
t.
only three
sible.
suppositionsare admis-
In
this
If
r"t\
case
sides
two
of
unequal,the
Zi2"Zr.
then
sides
the
equal.
opposite are
Given
angles of
two
the
Z
a
Z
are
opposite
greater side
oppositethe
A
"
the
less side.
"156.
3.
This
4.
Next
is
5.
This
6.
.-. r
r"t;
suppose
Z.R"
then
is
By hyp.,Zs
impossible.
Same
reason
as
2.
impossible.
Same
reason
as
3.
The
Q.E.D.
Ex.
an
Cor.
155.
isosceles
An
The
triangle.
"
r
proved
equiangular triangle is
bisectors of the base
r "
suppositions,
two
and
163.
ZT.
Zr.
t.
=
=
angles of
an
t, have
t
been
false.
also
equilateral.
isosceles
triangleform
58
PLANE
Ex.
perimeter
Ex.
of the
greater
Ex.158.
angles A
Ex.
of the
sum
the
than
B
on
the
than
and
intersect at 0, prove
BD
from
with
AO
Prove
Prop.
Ex.
162.
Upon
a
"
XVI
given
If the four
The
164.
to each
other
the
bisectors
Let
of
sides of the
triangleson
and
BD
AD
the
triangleto any
triangle.
base
same
BC, and if AC
=
and
and
isosceles.
a
bisector
Find
triangleone
of
the
of Ex.
69.
of the base
the.largerbase angle
locus of P.
quadrilateralare
equal, its diagonals
are
equilateralquadrilateral
an
bisect
they
the
angles of the
are
dicular
perpen-
quadrilateral.
is the
one
bisector
and
the
prove
the
of
of
correctness
answer.
Ex.
the
a
rightangles
adjacent sides of a quadrilateralare equal and
equal, one
diagonal is the perpendicular bisector
Tell which
other.
your
are
BC.
isosceles
an
is constructed
base
sides of
other, and
sides
two
D
AB.
by using the figureand method
diagonals of
If two
165.
Ex.
and
B
other.
bisect each
Ex.
"
of
two
are
The
angles of which is double the other.
the opposite side at the point P.
meets
163.
the
BO.
equal
triangleAOB
161.
AG
"
of the
one
with
vertex
that AC
Ex.
Ex.
triangle is less than
greater than
the
ABB
side of it such
same
AD
Prove
line drawn
If ABC
160.
0.
at
point in the base is less
Ex.
Prove
triangle ABC,
meet
A
159.
of any
quadrilateralABCD
CD.
Given
and
altitudes
triangle.
Given
157.
BC
and
The
156.
GEOMETRY
If,from
166.
a
in
point
a
perpendicular
cutting off equal segments from
oblique lines are equal.
drawn
Ex.
State and
167.
prove
the
converse
to
a
line,oblique lines
the foot of the
of Ex.
perpendicular,
166.
If,from a point in a perpendicularto a line,obliquelines are
drawn
cutting off unequal segments from the foot of the perpendicular,
the oblique lines are
unequal. Prove by laying off the less segment upon
Ex.
the
168.
greater.
Then
Ex.
use
166, " 153, and " 164.
point in a perpendicular to a line,two unequal
oblique lines are drawn to the line,the oblique lines will cut off unequal
segments from the foot of the perpendicular. Prove by the indirect method.
Ex.
169.
Ex.
170.
anglesof
Ex.
a
171.
If,from
any
that the
of Prop. XIV, prove
By means
triangleis less than two rightangles.
Construct
the altitude to the third
a
triangleABC,
side,c.
given
(See " 152.)
two
sum
sides,a
of
and
any
two
6, and
XVIII.
Proposition
The
167.
than
the third
A
Given
To
prove
168.
and
of
sum
Cor.
Cor.
two
sides
Theorem
of
a
triangle
is
greater
side
ABC.
a
+
I.
c
II.
the.partsof a
"
Any
greater than
169.
any
59
I
BOOK
b.
side
the
Any
broken
of a triangle is less than
differenceof the other two.
straight line is less
line having the same
than
the
the
sum
extremities.
sum
of
60
PLANE
GEOMETRY
that
preferto assume
line is the shortest line between
two
a straight
points may
omit Prop. XVIII
entirely.Then Prop. XVII may be proved
by a method similar to that used in Prop. XV.
(See Ex. 172.)
170.
Note
Teachers
Teacher.
to
who
"
"
Ex.
172.
Prove
Ex.
173.
If two
values
Ex.
If the
174.
Ex.
be
as
two
non-intersectingline
is greater than
the
ments
seg-
sum
of
PO
a
passing
point O,
shall
OB
+
possible.
illustrates the
law
light is reflected from a
light from the object,
L,
as
the
extremities
less than
the
sum
Note
to
P
as
of the
other
be
the
BG
is in front of it.
point within
any
side of the
until it intersects
171.
come
mirror
of any
ABC
Let
Hint.
the
If from
176.
to
appears
far behind
Ex.
AD
not
find
that
such
ing
limit-
points, P
two
To
is reflected and
from
what
9, between
The
mirror.
to
either.
which
P,
of any
joining lines
of the
sum
line AB
exercise
This
by
a
small
as
" 158.
in
ft
Given
175.
through
on
AB,
contained
lines.
two
R, and
and
14 and
triangleare
a
opposite ends
joined,the
other
the
sides of
hint
the third side be ?
must
are
by using the
XVII
Prop.
two
triangle lines
a
triangle,the
sides
of the
sum
Up
to this
been
complete,including argument
work
it is
of these
drawn
lines is
triangle.
given triangle,D the point
at E.
Apply Prop. XVIII.
Teacher.
are
within.
Prolong
pointall proofsgiven have
and
In
reasons.
frequently convenient, however,
to
have
written
students
give the argument only. These two forms will be distinguished
and the latter,
by callingthe former a complete demonstration
which is illustrated in Prop. XIX, argument
only.
It is often a sufficient test of a student's understanding of a
theorem
a
to have
proof.
XXXV,
form
or
This
him
may
state
be
in the form
merely the
given in
of
will be called outline
a
main
enumerated
as
paragraph,
of
proof.
in
points involved in
steps,as in Prop.
This
Prop. XLIV.
XIX.
Proposition
the
triangles have
Theorem
sides
of
equal respectively
sides of the other, but the included
to two
angle of
angle of the second, then
firstgreater than the included
the third
side of
third
side of the firstis greater than
If
172.
the
61
I
BOOK
two
two
one
the second.
and
ABC
DEF
with
AC
prove
"
2.
Place
A
DEF
AB,
D
upon
its
Draw
BK
4.
Draw
KF.
5.
In A
6.
.'.
upon
=
KBC,
DEy
BO
EF,
=
but
Only
that
BE
shall fall upon
its
equal
B.
AB
and
Denote
BC.
A
DEF
in
ABF.
and
FBC
BC
=
meeting
AC
at K.
BF.
BK.
Zkbc.
A
=
A
KBC.
.'.KF=KC.
10.
AK+
11.
.'.
12.
That
13.
and
=
9.
E
so
fall between
Zfbk
FBK
ABC
bisectingZ
FBK
BK
A,
A
positionby
new
3.
8.
on
will then
EF
7.
=
DF.
Argument
1.
AB
/-F.
/.ABC"
To
A
two
Given
..'.
KF"AF.
AK-\-KC"AF.
is,AC"
AC"DF.
AF.
Q.E.D.
PLANE
62
(a) Draw
177.
Ex.
falls
when
F
Prop.
XIX,
greater
taking AC
Prove
In A A
Hint.
;. Z. CEA
point
P
be
179
be
=
Q
XIX
(") Discuss
and
angle CAB
A.
A BE
"
and
~
by using Fig. 1.
i
FlG.
triangleABC
on
with
point Q
the conclusion
taken
were
CB
on
In
angle DAB,
Prop. XIX
on
AB
greater
CB,
so
than
that
AP
and
BC,
let
Prove
CQ.
"
CP.
if P
and
CD,
AB=DE,
DF,
=
for
case
triangleABC.
"ZEAC.
AB
Would
181.
Ex.
AB
taken
true
longed and
Prop.
Given
180.
Ex.
(the base)
CE, A CEA
greater than
^4^
falls within
F
the
B
Z. EAB
"
179.
Ex.
discuss
angle D.
than
178.
Ex.
figure and
a
when
AC;
on
GEOMETRY
AB
on
prolonged
of Ex.
?
Prove.
quadrilateral ABCD
angle OZM
AC
prove
is greater
greater
B
pro-
than
Proposition
if
than
BD.
XX.
Theorem
(Converseof Prop. XIX)
If two triangles have
173.
sides
to two
firstgreater
of
than
the
the
sides
two
other,
third
side
but
of one
the
equal
third
side
tively
respec-
of
of the second, then
the
the
angle opposite the third side of the firstis greater than
the angls opposite the third side of the second.
C
Given
To
prove
A
ABC
/-B
and
"
DEF
Z.E.
with
AB
D
=
DE, BC
=
EF, but
AC
"
DF.
63
I
BOOK
REASON8
Argument
1. Zb"Ze,
1. In this
Ze,
=
only three
case
Zb"Ze.
or
2.
Zb
are
Zb
First suppose
then
AC
positions
supadmissible.
2.
"ZE\
A
If two
"DF.
have
sides of
two
to
equal respectively
sides of the other,
one
two
but
included
the
the
the first "
Z of the
Z
of
included
second, then the
third side of the first "
the
third
second.
3.
This
4.
Next
is
impossible.
Zb
suppose
then
AABC
=
=
Ze
;
By hyp.,AC
4.
Two
DF.
"
equal
are
one
the
sides
included
other.
5.
AC"BF.
This
7.
.-.
is
impossible.
Homol.
Zb"Ze.
6.
Same
7.
The
Ex.
diagonals are
Ex.
the
183.
third
Ex.
drawn
to
Ex.
tw",
the
If two
sides of
triangleare
unequal angles with
If from
point
angle BSP
3.
Zb
suppositions,
ZE
Zb
and
been
P
in the
If
one
trianglecan
base
Ze,
=
proved
false.
S of
BT
so
an
isosceles
HP
median
drawn
to
side.
the third
that
the
triangleEST
is greater than
a
line is
PZ7,
then
angle PST.
angle of
be
unequal,
a
the vertex
is greater than
185.
as
110.
opposite sides of a quadrilateralare equal,but its
unequal, then one
angle opposite the* greater diagonal is
one
angle opposite the less diagonal.
side makes
184.
the
If two
182.
greater than
of
equal. "
reason
have
Q.E.D.
Z
and
ures
parts of equal fig-
two
"
Z
" 107.
are
6.
if two
equal respectively
are
to two
.-.
the
sides and the included
ADEF.
of
5.
of
" 172.
3.
A
side
divided
a
triangleis equal
into two
isosceles
to
the
sum
triangles.
of the other
64
PLANE
SUMMARY
OP
(a)
174.
If two
GEOMETRY
THEOREMS
When
the
sides of
a
FOR
triangleare
greater side is greater than
When
(6)
the
side of the
third
side of the second.
(c)An
OF
The
FOR
the lines
of any
sum
the
in the
are
first greater
angle opposite the
the
angle opposite the
than
LINES
PROVING
sides of
two
side of the
triangleis greater
a
THEOREMS
(a) When
175.
third
greater than
angle of
angle.
SUMMARY
third
first is
to
equal respectively
one
second,then
exterior
interior
sides of
the
than the third side of the
third
differenttriangles:
two
other,but
sides of the
two
in
:
triangle
same
unequal,the angle opposite the
angle oppositethe less side.
the
angles are
triangleshave
If two
in the
angles are
UNEQUAL
ANGLES
PROVING
either
UNEQUAL
triangle:
same
triangleis greater
a
mote
re-
than
the
side.
side of
Any
the difference
a
triangleis
of the other
less than
the
sum
and
greater than
two^
angles of a triangleare unequal,the side oppositethe
greater angle is greater than the side oppositethe less angle.
If two
(b) When
the lines
in
are
differenttriangles:
triangleshave two
sides of the other, but
If two
two
sides of
the
included
greater than
the
included
third
the
first is greater than
side
of
to
equal respectively
one
the
angle of
the
angle of
second,
the
first
then
the
side
third
of
the
a
line
second.
(c)Every point not
is not
a
perpendicularis
of
of the line.
the shortest
line
straight
from
a
point
line.
Any
broken
In
perpendicularbisector
the ends
equidistantfrom
The
to
in the
some
straightline is less
line having the same
texts
Exs.
168
and
169
than
the
sum
of
extremities.
are
given
as
theorems.
the
parts of
a
66
PLANE
GEOMETRY
Proposition
180.
If
XXI.
straight lines
two
straight line, they are
Theorem
parallel
are
parallel to
each
to
a
third
other.
D
lines
Given
To
181.
182.
of the
b, each
IIc.
IIb.
A
Def.
other
more
a
prove
and
a
transversal
is
a
line
that
intersects
two
or
lines.
straightlines
eightangles formed,
Defs.
3, 4, 5, 6 are
1, 2, 7, 8 are
4 and
5, 3
If
two
interior
exterior
and
6,
are
cut
by
a
transversal,
angles;
angles;
are
alternate
angles;
alternate
1 and 8, 2 and 7, are
exterior
angles;
1 and 5, 3 and 7, 2 and 6, 4 and
also
8, are corresponding angles (called
interior
exterior
interior
angles)
Proposition
making
lines
XXII.
straight lines
a
pair of alternate
parallel.
If
183.
are
67
I
BOOK
two
Theorem
cut
are
by
interior
transversal
a
angles equal, the
N
M
"*P
Q
B"
Given
two
points C
and
To
str. lines MN
D,
Z
MOD
OQ
=
the transversal
by
cut
ZQDC.
Argument
1.
and
MN
2.
3.
OQ
Reasons
either IIor
are
this
In
only two
case
are
Suppose that if JV is not IIOQ;
then
they will meet at
some
point as P, forming,
with
line DC, A
Then
1.
II.
not
in
AB
IIOQ.
MN
prove
making
and
/.MOD
"
2.
suppositions
admissible.
def. of IIlines.
By
" 177.
PDC.
ZQDC.
3. If
side
one
of
a
A
is prolonged,
the ext. Z formed
"
either
int. A.
4.
This
5.
/.
MN
is
Z.MCD
impossible.
The
IIOQ.
and
Q.E.D.
184.
Cor.
If
I.
=
of
the
remote
" 153.
ZQDC, by hyp.
supposition that
OQ
are
not
IIhas
Prove
a
been
proved false.
versal
straight lines are cut by a transpair of corresponding angles equal, the
two
making a
lines are
parallel.
Bint.
MN
pair of alt. int. A equal, and
apply the theorem.
68
PLANE
GEOMETRY
versal
If two straight lines are cut by a transmaking a pair of alternate exterior angles e-qual,
the lines are parallel. (Hint. Prove a pair of alt. int. A equal.)
185.
Cor.
186.
Cor.
same
lines
If
versal
straight lines are cut by a transmaking the sum
of the two interior angles on the
side of the transversal
equal to two right angles, the
are
parallel.
^
HI.
lines
Given
transversal
and
m
+ Z2
n
cut
the
by
J
m
=
2rt.A
jt
||n.
m
prove
two
making
t
Zl
To
II.
Reasons
Argument
1.
Zl
+ Z2
=
2.
Zl
+ Z3
=
2rt.A
1.
By hyp.
2rt.Zs.
2.
If
str. line
one
meets
other
an-
line,the sum
two
adj. A is
str.
of
the
rt. A.
3.
.". Zl
+ Z2
=
Zl-fZ3.
3.
Z2
4.
.-.
5.
.*. m
=
Z3.
4.
If
same
each
" 54, 1.
subtracted
equals are
from
ders
equals,the remainare
equal. " 54, 3.
5. If
IIn.
" 65.
Things equal to the
thing are equal to
other.
2
two
str. lines
are
cut
by a transversal making
a pairof alt. int. A equal,
the lines
Q.E.D.
187.
to
a
Cor.
third
190.
II.
" 183.
If two straight lines are perpendicular
straight line, they are parallel to each other.
IV.
straightlines are
of the two exterior angles on the same
rightangles,the lines are parallel.
Ex.
are
If two
cut
by
a
transversal
making
side of the transversal
the
sum
equal to two
I
BOOK
Ex.
BGE
the
In
191.
diagram, if angle
annexed
angle CHF,
=
are
69
and
AB
CD
"
parallel?
Trove.
V
_^"
A
In the
192.
Ex.
diagram,
same
if
angle HGB
Q
R
"
*""
angle GHC, prove that the bisectors of these
angles are parallel.
=
extremities
straightlines bisect each other,the lines joiningtheir
parallelin pairs.
are
the
In
194.
Ex.
supplementary,
are
two
prove
AB
Ex.
adjacent angles of
will
sides of the quadrilateral
Proposition
Through
188.
to
a
be
angle
FHD
tary,
supplemen-
quadrilateralare
any
parallel.
Problem
XXIII.
given point
a
and
191, if angle BGE
parallelto CD.
diagram for
If two
195.
Ex.
\.tt
If two
193.
Ex.
D
to construct
a
line
parallel
given line.
""Y
A
A
line
D
K
and
point P.
construct, through,point P,
Given
To
/R
AB
line IIAB,
a
I. Construction
1. Draw
a
2. With
P
ZBRP.
3.
line
through
and
vertex
as
P
cuttingAB
PS
as
at
some
point,as
side, construct
Z.
R.
YPS
=
"125.
XY
will be
The
II.
IIAB.
proof and
discussion
are
left
as
an
exercise for the
student.
Ex
196.
using : (a) "
Through
183 ;
a
(6) "
given point construct
185
;
(c) " 187.
a
parallelto
a
given line by
70
PLANE
GEOMETRY
Proposition
XXIV.
Theorem
(Converse of Prop. XXII)
If
189.
alternate
and
To
interior
||lines
Given
G
parallel lines
two
angles
and
AB
are
cut
CD
are
by
cut
equal.
by
the transversal
Z
prove
A GH
Either
Z
A GH
=
Z DHG,
that
ZXGH=
Then
XY
1. In
this
2.
XY,
makes
ZDHG.
3.
||CD.
With
given vertex and
a
given side,an Z may
be constructed
equal to a
" 125.
given Z.
If two str. lines are cut by
transversal
a
making a
pair of alt. int. A equal,
a
the lines
4.
But
5.
It is
||CD.
AB
and
impossiblethat
XY
.'.ZAGH=
both
are
AB
positions
sup-
admissible.
are
line
G,
only two
case
ZDHG.
through
6.
points
Reasons
Suppose ZAGH^ZDHG,
but
3.
at
Z DHG.
=
ZAGHJ=
or
2.
EF
H.
Argument
1.
transversal, the
a
4.
By hyp.
5.
Parallel
are
line
|| "183.
.
post.
" 178.
supposition
that
||CD.
ZDHG.
6. The
ZAGH^Z
Q.E.D.
proved
DHG
false.
has been
71 ^
I
BOOK
(Converse of Cor. I of Prop. XXII). // two
parallel lines are cut by a transversal, the corresponding
A*
angles are equal.
190.
I.
Cor.
||lines
two
Given
and
XY
by the transversal AB,
correspondingA 1 and
out
To
Z 3
Hint.
=
Z2.
A 2
by "
Z 1
prove
=
MN
ing
form2.
^
189.
\"
(Converse of Cor. II of Prop. XXII).
nate
If two parallel lines are cut by a transversal, the alterexterior
angles are equal.
191.
Cor.
n.
192.
Cor.
m.
If
two
of
the
parallel lines are cut
two interior angles on
is two
right angles.
two
Given
by
cut
IIlines XY
the transversal
int. A
To
of
(Converse
1 and
AB,
by
the
Ill
of
Prop. XXII).
transversal, the
a
side
same
of the
sum
versal
trans*
MN
X_
ing
form-
2.
jf
Z 1 + Z 2
prove
and
Cor.
=
2 rt. A.
Q.E.D.
193.
two
the
IV.
two
sum
Cor.
V.
straight line perpendicular
not
(Opposite
straight lines
of
the two
the transversal
are
A
parallelsis perpendicular
194.
If
Cor.
not
are
interior
equal
to the
of
Cor.
cut
by
Ill
a
angles
to two
other
of
to
the
Prop. XXII).
same
making
side
right angles, the
parallel. (Hint. Apply " 137,
or
use
the
of
also.
transversal
on
one
indirect
of
lines
method.)
72
PLANE
195.
Cor.
Two
VI.
GEOMETRY
lines
intersectinglines
two
and
CD, and
To
also intersect.
lines
intersecting
two
Given
perpendicular respectivelyto
_L AB,
BE
that
prove
"
CF
CD.
and
BE
A
AB
also
CF
intersect.
Argument
196.
they intersect
angles formed
seven
angles ?
its
angles will
Zfcd
is
3.
.-.
4.
Likewise
5.
.-.
Z1
+ Z2
6.
.-.
BE
and
or
a
a
Zl."
rt. Z.
art.Z.
Z2
"
rt. Z.
2rt.
"
A.
also intersect.
CF
lines
more
a
said
are
to
In
parallel to
CD,
two
of its sides
and
annexed
is
EF
one
of the
of the
other
that
each
"
diagram AB
parallel to
.
so
two
parallel,
supplementary.
the
if
concurrent
point.
common
quadrilateralhas
a
q.e.d.
be
parallelsare cut by a transversal
45", how
degrees are there in
many
be
199.
Ex.
2.
at
is
If
198.
Ex.
CB.
If two
197.
Ex.
1. Draw
Two
Def.
Only
is
pairs of
-,
%
OH.
Prove
(a) angle 1
(")
angle
If
200.
Ex.
"
1 +
a
angle,parallelto
\
angle 2
;
angle 3
=
2
line is drawn
of the sides
one
\S
F
rightangles.
H
through any point in the bisector of
of the angle,an isosceles trianglewill
an
be
formed.
sides
Draw
201.
Ex.
so
the
that
sum
a
line
parallelto
of the segments
the
base
adjacent to
of
a
the
triangle,
cutting the
base
shall
equal
the
parallelline.
Ex.
of
a
202.
pair of
If two
the
parallellines are cut by a transversal,
corresponding angles are parallel.
bisectors
74
GEOMETRY
PLANE
Two
198.
either
are
To
Z5
XXV.
angles whose
sides
equal
Zl
Given
Proposition
and
parallel,each
are
the A
at E, with
Z2=Z3,
II EF
AB
Zl
and
+
and
Z4
IIDE.
A C
2 rt. 4
=
Zl
+
2rt.Zs.
=
Reasons
Argument
1.
each,
to
supplementary.
or
Zl
prove
Theorem
Prolong
they
and
".4
Z"J"
intersect
point as
at
until
1.
"54,16.
Str.linepost.il.
some
H.
2.
Z1
=
Z6.
2.
lines
Corresponding A of ||
are
equal. " 190.
3.
Z2
=
Z6.
3.
Same
4.
.-.
Z1
4.
Things equal to the
thing are equal to
=
Z2.
reason
other.
5.
5.
Z3=Z2.
If two
2.
as
same
each
" 54, 1.
str. lines
the vertical A
intersect,
are
equal.
"77.
6
.-.
Z1
7. Z2
=
+
Z4
6.
Z3.
=
2rt.A
Same
7.. If
reason
one
str. line
str.
of the
rt. A.
8.
.-.
Z
1 +
Z 4
=
2 rt. Zs.
8.
4.
as
two
meets
line, the
adj.A
other
ansum
is 2
" 65.
Substituting Z
equal Z 2.
1
for
its
75
1
BOOK
Reasons
Argument
9. Z5=Z4.
10.
.-.
9.
Zl
Z5
+
2rt.A.
=
10.
annexed
side ; thus, in the
In the
Z4,
whose
ED)
are
EF)
sides
rightside of ZA
and
the
left,I.
r
Fig.
2,
and
of
Prop. XXV,
Zl
and
left to left
(AC
diagram
right(ABto
is
||rightto left (AB
are
Hence
supplementary.
has
the
diagram,
rightside of Z.B
side,I ; the
5.
as
Substituting Z5
its vertex
from
Every angle viewed
Note.
reason
its
for
equal Z4.
Q.E.D.
199.
Same
to
to
Z2, whose
EK)
EF)
are
and
right and
a
is
r
and
Fig.
3.
sides
are
equal ;
left to
a
left
the left
||rightto
while
and
Zl
right {AG
to
:
angleshave their sides parallelrightto right
and leftto left,
they are equal
(b) If two angles have their sides parallelrightto leftand left
to right,
they are supplementary.
(a) If
200.
Ex.
four
Fig.3, above, show
In
213.
two
angles are
Ex.
a
has
quadrilateral
Given
215.
other,are
216.
sides of
to the
side
th"
third
217.
two
the
If two
homologous
Ex.
rightside
of each
of the
its
its opposite
oppositesides parallel,
equal.
side of the
Ex.
If
214.
is the
O.
angles about
Ex.
which
equal angles having a side of one
?
other sides necessarilyparallel
of the
second.
Construct
a
drawn
'altitude
from
a
Prove.
triangle are parallelrespectivelyto two
the third side of the first is parallel
equal triangle,
sides of
an
parallelto
the
a
triangle,given an angle and
of the given angle.
vertex
its bisector
anil
76
PLANE
GEOMETRY
Proposition
201.
Two
each,
angles whose
either
are
XXVI.
equal
Theorem
sides
perpendicular, each
are
supplementary.
or
H
K
C
Zl
Given
To
Prove
OE
Draw
Z 7
202.
Z
=
IICK
1, and
right (OA to PIT)
and Z4, whose
sides
P^T),
to
rightand
drawn
=
seen
that
Z\
and
(OB
right to left
supplementary. Hence
two
angles
rt. A.
Z",Z1+Z5=2
Prove
"
are
J_ HD.
OB
OE
"
OA
and
OF"
OB.
Z 2.
left to left
and
and
EC
have
Z2,
whose
PIT), are
(0.4 to PC)
to
sides
equal ;
and
right
"
are
while
left to
Zl
right
:
their sides
perpendicularrightto
equal.
angles have their sides perpendicular rightto left
right,they are supplementary.
218.
to
Z 7
DH.
D
two
leftto
Ex.
OFW
leftto left,
they are
(b) If
and
are
(a) If
203.
and
It will be
Note.
_L
OA
Zl+Z4=2rt.
prove
to
(OB
at P, with
Z1=Z2=Z3,
prove
Hint.
the A
and
to
If from
a
the sides of the
point outside of an angle perpendiculars are
angle, an angle is formed which is equal to the
given angle.
from
the
right triangleif a perpendicular is drawn
is divided
of the right angle to the hypotenuse, the right angie
vertex
into two angles which
are
equal respectivelyto the acute angles of the
Ex.
219.
In
a
triangle.
Ex.
220.
If from
the
end
of the
bisector
of the vertex
trianglea perpendicular is dropped upon
perpendicular forms with the base an angle equal to
isosceles
one
of
of
angle
the
arms,
half the vertex
an
the
angle.
BOOK
XXVII.
Proposition
The
204.
Theorem
of the angles of
sum
77
I
any
triangle is
two
right
angles.
A
Given
To
ABC.
A
prove
A
+ AC
/.ABC
+
2 rt. A.
=
Reasons
Argument
Through
Al
draw
B
A 2
/ABC+
+
AC.
BE
1. Parallel
line
2. The
of all the ^
2 rt. A
=
a
sum
point on
post.
"179.
about
side of
one
a
str. line
passing through
2
rt. A.
point
that
=
"66.
3.
Zl=Zi.
3. Alt.
int. A
equal. "
of
II lines
are
189.
4.
Same
5.
Substitutingfor A 1 and 2
their equals,z" ^1 and
C,
reason
as
3.
respectively.
205.
are
Cor.
I.
In
right triangle the
a
two
acute
angles
complementary.
206.
angle
207.
Cor.
or
one
Cor.
II.
In
208.
Cor.
angle of
othgr acute
one
right
angles of one
triangle are
angles of another, then the third
equal
IV.
can
angle.
If
to two
respectively
of the firstis equal
be but
one
obtuse
in.
triangle there
a
If
two
to the third
two
equal
angles are
to
angle of
angle
the second.
acute
an
right triangles have
acute
an
angle of the other, the
equal.
78
PLANE
209.
Cor.
Two
V.
and
to the
and
an
and
211.
VII.
hypotenuse
hypotenuse
and
a
and
ABC
Prove
Cor.
right triangles are
equal if
side of one
to
are
equal respectively
side of tJie other.
a
A AB
prove
212.
C
A
=
DFK
a
is
Cor.
third
one
214.
not
of each
Ex.
angle
Ex.
twice
Ex.
an
two
V
221.
of the
222.
If any
If the
223.
Find
If
vertex
the
first and
225.
and
A DEK
upon
BC=EF.
is isosceles.
the base of an
also the vertex
is the word
angle of
isosceles
angle.
equilateral triangle
60".
homologous
used
isosceles
in
Cor.
VI
but
triangleis 60",what
If the
of
of the base
angle of a triangleis
third angle.
base
value
isosceles
an
triangleis 20", find
angles.
angle of a triangleif the second
third equals three times the second.
angle of
angle.
vertex
is the
angles?
angle
bisectors
the
one
expressionfor each
an
each
expression for the
Ex.
altitude
remaining
two
by
224.
and
BE
the
?
included
the
=
angle of an
right angles, or
Why
Question.
in Cor.
Ex.
of
AB
the
BEF.
base
Each
IX.
with
BEFy
str. line ; then
The
VIII.
triangle bisects the
213.
are
Two
and
rt. A
Given
Hint.
potenu
hy-
equal respectively
angle of the other.
acute
an
the
Two
VI.
Cor.
To
and
equal if
are
right triangles are equal if a side
acute
angle of one
are
equal respectively to a
the homologous acute
angle of the other.
Cor.
an
side
right triangles
acute
angle of one
hypotenuse
210.
GEOMETRY
an
m"
and
isosceles
another
the
angle equals
angle Z",write
triangleis a", write
an
BOOK
Ex.
226.
Construct
Ex.
227.
Construct
How
large is the
I
angle of 60"
an
79
; 120" ; 30" ; 15".
righttrianglehaving
acute
angle ?
a
other
Ex.
228.
Construct
Ex.
229.
Prove
Prop. XXVII
Ex.
230.
Given
two
Ex.
231.
Find
Ex.
232.
The
of its acute
one
angles60".
angle of 160".
an
by using each
of the
diagrams given
above.
a
the
angle
The
233.
to
of two
to each
half the
234.
If two
235.
If in
the
angles of
the
bisectors
.sum
of the
of two
two
adjacent angles of
remaining angles.
the bisectors of the base
angle
angle.
third
quadrilateral.
a
the
exterior
the
angle at
isosceles
an
the
vertex,
one
of its ends
the
opposite side prolonged
will form
side and
of
triangle,construct
a
formed
angles of
an
celes
isos-
by prolonging the base.
straightlines are cut by a transversal and the bisectors
interior angles on the same
side of the transversal are perpendicular
other,the lines are parallel.
Ex.
fourth
between
angle between
triangleis equal to
Ex.
of the
sura
quadrilateralis equal
Ex.
angles of
and
with
the
the
the
line drawn
of the
base
angles is
perpendicular to
the
base
one
at
meeting
adjacent
exterior
opposite
a
triangle each
portion
side
an
equilateraltriangle.
whose
Two
236.
Ex.
sides
angles
perpendicular each
are
to
(Prove by using the annexed
Ex.
If at the ends
237.
of the
each
either
are
equal
or
mentary.
supple-
diagram.)
hypotenuse of
a
right triangleperpendiculars
meeting the other two sides of
hypotenuse are drawn
triangleprolonged, then the figurecontains five triangleswhich are
to
the
the
mutually equiangular.
Ex.
find
the
two
If
238.
the
other
triangle.
remote
one
angle of
interior
What
interior
a
angle of
relation
angles of
triangleis 50" and
the
is there
the
triangle; also
between
triangle?
an
another
the
angle is 70",
exterior
exterior
angles of
angle and
the
80
PLANE
GEOMETRY
Proposition
An
215.
of the
two
XXVIII.
Theorem
exterior
angle of a triangle is equal
interior angles.
remote
to the
sum
vertex
of
B
C
A
Given
To
The
Draw
239.
isosceles
Ex.
three
BOB
proof is
Hint.
Ex.
A
prove
A
=
left
CE
as
-f AB.
A
exercise for the student.
an
The
bisector
of
triangleis parallelto
240.
241.
If the
The
exterior Z.
an
IIAB.
the
of the
the
at
an
base.
angles of
righttriangle.
exterior
triangleis
sum
angle
exterior
an
of two
sura
rightangles,the
Ex.
Z.DCB
with
ABC
a
three
exterior
a
triangleis equal
angles of
triangleis
a
to
four
rightangles.
Ex.
242.
What
is the
Ex.
243.
If the
two
of the
sum
exterior
exterior
angles at
these bisectors
bisected,the angle between
base
the
interior
angles of
triangle.
Ex.
244.
If BE
and
bisects
triangleABC,
angle DA C, angle E
angle C.
bisects
AE
is
equal
angle
B
245.
D
to
one
246.
The
equidistantfrom
angle either acute
is
base
equal to
?
quadrilateral
of any
half the
triangleare
sum
of the
of
half
is any
=
Ex.
the
a
the exterior
point in the base
BC
of isosceles triangleABC.
The side
AC
is prolonged through C to E so that
CE
is drawn
CD, and DE
meeting
equal to three times angle AEF.
Ex.
angles of
mid-point
the
three
angle.
of
the
vertices.
AB
at
F.
Prove
angle
EFA
hypotenuse of a right triangle is
Prove
by laying off on the right
82
PLANE
GEOMETRY
Proposition
218.
the sides
If
one
thus
the
formed
is
sion
prolonged in successides
being prolonged
of the exterior angles
adjacent
vertex, the
same
Theorem
polygon are
any
two
no
way,
through
of
XXX.
sum,
four right angles.
4/D
Given
To
with
P
polygon
angles.
Z 2 + Z3
Z 1 +
prove
Zl, Z2, Z3, Z4,
+
Z4
+
..-its successive
4 rt. A.
=
"""
Argument
1.
Zl+^6
Z7
.-.
the
3.
n
of the
the
then
str. line
one
the two
one
" 65.
n
2.
+
=
Z2
by
J
E+I=2
Z;
4
rt.
+
Z3+Z4+
4rt.Z.
of
Z.
3.
and
n
4.
The
the
by
rt. Z.
sum
-
=
line,the sum
adj.Z is 2 rt.
If
any
.-.."
other
an-
equals are multipliedby
equals,the products are
equal. " 54, 7 a.
Arg. 2.
vertices
4. But/=2nrt.Z-4rt.Z.
5.
meets
str.
int.
int. and
of all the ext. A
sum
If
;
of all the
sum
interior A
E\
at
on
rt. A.
the
Denote
of the
1.
+
2 rt. A,
=
sum
2
so
the ext. Z
ext. zs at
=
and
Z,
sum
and
vertex
2.
Z2
4
rt.
2 rt.
=
i.e. the
Z
Reasons
2
=
terior
ex-
t\e. Zl
...
q.e.d.
5.
If
of all the
polygon
4 rt. Z.
=
2
n
Z
of
rt. A
" 216.
subtracted
equals are
from equals,the remainders
are
equal. " 54, 3.
219.
form
useful iu the
more
formula
The
Note.
"fix. 247.
Find
(;i
rt. A
n
of the
sum
4 rt. A
"
(" 216)
is sometimes
2 rt. A.
2)
"
the
2
83
I
BOOK
angles of
a
polygon
7 sides ; of 8
of
sides ; of 10 sides.
Prove
248.
Ex.
possiblefrom
XXIX
by drawing
How
diagonals
many
polygon
of diagonals possible in
sides ?
n
of
polygon
a
from
drawn
be
diagonals can
many
of 8 sides ? of 50 sides ? of
n(n
as
Show
that
sides
n
in
vertex
one
the greatest
(using
all
a
ber
num-
vertices)is
3)
-
as
vertex.
one
249.
Ex.
Prop.
,
2
Ex.
How
250.
degrees are
angle of
many
? in each
quadrilateral
Ex.
14
How
251.
rightangles ?
Ex.
angles is double
Ex.
253.
to
be
Ex.
it
angles equals 12"
255.
Ex.
angles is
Ex.
one
256.
angles is
Ex.
How
258.
make
with
a
a
sides
many
sides
How
the
of whose
sum
angular
equi-
an
?
angles is
540" ?
a
exterior
an
angle
has
a
sum
of
an
each
polygon
How
Tell
the
sum
many
sides has
of its
a
polygon
what
How
to
each
of
whose
interior
gon
equiangular poly-
of whose
exterior
a
an
of
whose
exterior
angle ?
polygon the
exterior
angles?
sides has
angles is
vertex
has
of its adjacent interior
many
pavement.
common
of
144"?
many
eleventh
three of its exterior
Ex.
angle
equiangular pentagon
polygon the
exterior angles ?
possiblefor
140"?
each
?
six times
257.
of its
an
in
polygon
a
sides has
sum
72"?
How
has
rightangles ?
many
the
Is
70"?
254.
20
How
252.
sides
many
there
sum
of whose
equiangular polygon
if the
interior
sum
of
180" ?
be put together to
equiangular polygons can
be placed
equiangular triangles must
many
fillthe angular magnitude around
a point ?
84
PLANE
GEOMETRY
PARALLELOGRAMS
QUADRILATERALS.
QUADRILATERALS
220.
A
Def.
sides
221.
CLASSIFIED
its sides
PARALLELISM
TO
quadrilateralwhose
a
site
oppo-
parallel.
are
A
Def.
trapezoid
A
Def.
is
quadrilateral
having
a
the other
is
trapezium
a
two
of its
two
parallel.
not
having
quadrilateral
of
two
no
parallel.
PARALLELOGRAMS
223.
is
parallelogram
oppositesides paralleland
222.
RESPECT
WITH
CLASSIFIED
A
Def.
rectangle
RESPECT
WITH
is
a
ANGLES
TO
having
parallelogram
angle.
right
one
\
It is shown
later that all the
angles of
a
rectangleare right
angles.
224.
A
Def.
is
rhomboid
a
having
parallelogram
an
oblique
angle.
It
is shown
later that
all the
angles
of
rhomboid
a
are
oblique.
225.
a
a
A
Def.
rectanglehaving
adjacentsides equal is
two
square.
It is shown
later that all the sides of
a
226.
A
adjacentsides equal is
Def.
rhomboid
having
two
equal.
are
square
rhombus.
It is shown
227.
equal is
228.
later that all the sides of
A
Def.
an
isosceles
point in
229.
and
the
the
side of
its altitude
oppositeside
is
equal.
non-parallelsides
its two
parallelogrammay
perpendicular to
a
line drawn
The
Def.
are
trapezoid.
Any
Def.
its base, and
any
trapezoidhaving
rhombus
a
bases
a
of
a
line drawn
to the other.
is then
the
any
regarded as
the
base
from
altitude.
trapezoidare
from
be
its
parallelsides,
point in
one
base
pendicula
per-
BOOK
Proposition
Any
230.
equal, and
To
any
O
Given
two
85
I
XXXI.
Theorem
opposite angles of a parallelogram are
two consecutive
angles are supplementary.
ABCD.
prove
(a)ZA
(6)any
:
Zc, and
=
ZB
Zd
=
consecutive
two
;
A,
as
Argument
1. ZA
Z
=
C
and
and
A
B, sup.
Reasons
Zb=Zd.
If two
A
their sides
have
right and left
to left,they are
equal.
" 200, a.
IIlines are cut by a
two
IIrightto
2. A
and
A
B
are
If
sup.
transversal,the
the two
int. A
on
of
sum
the
same
side of the transversal
two
Q.E.D.
231.
Ex.
is
a
" 192.
All the
angles of a rectangle are right angles,
the angles of a rhomboid
are
oblique angles.
Cor.
all
and
rt. A.
is
259.
If the
oppositeanglesof
a
quadrilateralare
equal,the figure
parallelogram.
Ex.
260.
If
an
of
angle
one
another, the remaining angles are
Ex.
op
261.
a
The
bisectors
square) inclose
a
of the
parallelogram is equal
equal
each
angles of
rectangle.
to
an
angle
of
to each.
a
parallelogram (not a
bus
rhom-
86
PLANE
GEOMETRY
XXXII.
Proposition
O
Given
To
opposite sides
The
232.
The
Cor.
all the
234.
Cor.
left
as
All
I.
sides
and
CD
=
proof is
233.
235.
Cor.
of
Cor.
it into
rhombus
a
the
from
Ex.
so
A
rv.
a
263.
are
the
equal, and
are
square
equal.
intercepted between
lines
the
equal.
are
The
III.
of
are
Parallel
II.
for the student.
perpendiculars
points
two
any
diagonal of
drawn
to
one
of
in the other
are
equal.
parallelogram
a
two
divides
equal triangles.
perpendicularsdrawn
opposite vertices are equal.
262.
Ex.
and
two
exercise
the sides
parallel lines from
236.
equal.
are
BC=AD.
an
parallel lines
same
parallelogram
a
ABCD.
AB
prove
of
Theorem
to
The
The
diagonals of
diagonals of
a
a
rhombus
a
are
diagonal of
a
parallelogram
perpendicularto each other
square.
Ex.
264.
The
diagonals of
a
rectangleare
Ex.
265.
The
diagonals of
a
rhomboid
Ex.
266.
If the
diagonalsof
a
parallelogramare
equal,the figureis a
If the
diagonals of
a
parallelogram are
not
Draw
a
are
equal.
unequal.
rectangle.
Ex.
is
a
267.
equal,the figure
rhomboid.
Ex.
268.
line
portion interceptedbetween
Ex.
269.
Has
Explain
this been
the
the
parallelto
the
sides may
statement
proved ?
:
base
be
of
a
triangleso
equal to
Parallel lines
are
a
that
the
given line.
everywhere equidistant.
270.
Ex.
locus of
Find
the
Find
the
87
I
BOOK
point that
a
equidistantfrom
is
given
two
lines.
parallel
271.
Ex.
horizontal
line ;
vertical line ;
given
to OX.
Find
rightof
the
237.
the
one
lines in
inches
a
right of
to the
inch
one
and
line OX
given
a
line.
a
given line.
left of the
to the
point three
a
Rene" Descartes
Note.
if its
line OY
above
OX
perpendicular
and
inches
two
the fact that
of
the
the first to
(1596-1650) was
position of
a
point
in
a
plane
distances,
y, from
the
inch
horizontal
a
importance
and
x
(6)
point : (a)
a
above
inch
one
a
OY.
is determined
say
locus of
Historical
observe
point: (a)
below the given
the
locus of
the
of
inches
two
Given
273.
Ex.
to
Find
272.
Ex.
(")
locus
fixed
two
plane, perpendicular
other,are known.
He showed
that geometric figures
be represented by
can
veloped
algebraicequations, and deeach
to
subjectof analytic
the
his
name
Cartesian
as
Descartes
Tours
in
famous
France,
Fl"che.
He
and
most
sent
was
age
the
school
at
La
of
was
good
that
ily,
famDescartes
time,
of
latter,and
which
near
to
since, at
men
One
try.
geome-
and
of
Jesuit
by
born
was
eight years
at
is known
which
geometry,
positionentered either the
joined the army of the Prince
day, while walking
challenged every
problem.
in
a
street
read
who
one
in
church
of
a
or
the army,
Orange.
Holland
it to
town,
solve
a
he
a
short
twenty
time
years,
in
Paris, but
devoting his
and
physics.
His
him
the
of the
name
solved
work
in
Father
it with
soon
little
retired
to
Holland,
a
saw
where
placard
geometric
certain
is
and
difficulty
realized then that he had no taste for militarylife. He soon
commission
and spent five years in travel and study. After
Descartes
the
he chose
said
to
have
resigned his
this he lived
he
lived
for
philosophy, astronomy,
importance as to give
philosophy was
of Modern
Philosophy.
time
to
mathematics,
of
such
88
PLANE
GEOMETRY
Proposition
The
238.
XXXIII.
diagonals
of
Theorem
bisect
parallelogram
a
each
other.
O
Given
To
prove
Hint.
the
with
AO
OC
=
Prove
A
OBC
its diagonalsAC
and
BO
A
OB
=
274.
bisected
by
Ex.
at
intersecting
BB
0.
OB.
=
A.
AO
..
and
OC
=
BO
=
OD.
the
the
and
of the
sides
How
Ex.
If the
278.
diagonals
a
lines drawn
equal.
Ex.
Find
279.
of
diagonals of
on
parallelto
a
parallelogram
a
rhombus
parallelogram,the figureis
Ex.
point
having
a
of intersection
given
of the
parallelograms?
If the
277.
locus of the
is the
What
altitude ?
a
be constructed
parallelogramscan
many
other, the figureis
each
parallelogramand passing
of its diagonals is bisected at that point.
sides of
the
by
point of intersection
diagonals of all these
Ex.
given triangle.
line terminated
A
276.
Ex.
base
If
275.
through
are
and
parallel
through the vertices of a trianglelines are drawn
the opposite sides of the triangle,
the lines which
join the vertices of
to the
trianglethus formed
oppositevertices of the given triangleare
Ex.
to
A BCD
or
rhombus
side of
the
other
or
a
the
two
angles of
the
square.
trianglethe point from
a
to
square.
parallelogram bisect
a
one
a
perpendicular
are
sides,and terminated
which
by
straight
sides,
those
(See " 232.)
280.
Find
the
Find
the
locus of
a
given distance^ from
a
given
given distance from
ends of another given line.
a
given
point
at
a
finite line AB.
Ex.
line and
Ex.
281.
locus of
also equidistantfrom
282.
Construct
the altitude upon
the
a
the
a
point
at
a
parallelogram, given
given side.
a
side, a diagonal, and
90
PLANE
Ex.
284.
Construct
a
rectangle,given the
Ex.
285.
Construct
a
square,
Ex.
286.
by
of
means
Ex.
Construct
287.
Ex.
given
Through a given point
Prop. XXXIV.
(1) using "" 239
as
GEOMETRY
the median
drawn
of
238 ;
and
angle of
An
288.
median
a
a
and
base
side.
construct
parallelto
a
triangleby
a
altitude.
the
(2) using ""
220
of
means
and
a
a
given
line
gram,
parallelo-
238.
triangleis right,acute, or obtuse according
its vertex
is equal to, greater than, or less than
a
from
half the side it bisects.
Proposition
XXXV.
Theorem
If two opposite sides of a quadrilateral
parallel,the figure is a parallelogram.
240.
and
A
To
D
quadrilateral
ABCD,
Given
ABCD
prove
a
are
BC
of
Proof
Ex.
O.
Ex.
common,
1.
Draw
diagonal
2.
Prove
A
3.
Then
4.
.'.ABCD
a
BCD
A
"
Zcdb
a
ABD.
/.ABD
==
is
BD.
and
IICD.
AB
O.
parallelogram,having given a base, an adjacent
altitude,making your construction depend upon " 240.
Construct
the
291.
line
IIto AD.
mid-points of two opposite sides of a parallelogram
a parallelogram will be formed.
pair of opposite vertices,
290.
angle, and
equal and
If the
289.
joined to
Ex.
both
with
Outline
Ex.
equal
are
are
292.
the
If the
perpendicularsto
equal, then
If two
a
the
lines
are
a
other
any
two
points
vertices
and
a
in
other
an-
parallel.
parallelograms have
lines joiningthe
line from
two
four vertices
form
a
diagonal in
parallelogram.
Proposition
91
I
BOOK
Theorem
XXXVI.
(Converseof Prop. XXXIII)
If
241.
other, the
bisect
diagonals of a quadrilateral
figure is a parallelogram.
the
each
B
D
A
at
To
with
ABCD
quadrilateral
Given
0
so
ABCD
prove
that
AO
and
CO
=
its
1.
In A
and
OBC
AO.
3.
Z3
=
Z4.
.'.
.*.
a
YD
Ex.
Ex.
and
C
to
If each
AD.
is
and
X
a
O.
q.e.d.
let
diagonal AC
Y, respectively,making AX
half of each
diagonal of
joiningthe points of bisection
295.
Lines
296.
drawn
297.
between them.
from
form
the vertices
opposite sides
State four
conclusion,"the
Ex.
BC\\
parallelogram ABCD,
terminating in the
Ex.
ABCD
Z2.
=
prolonged
be
=
CY.
Prove
parallelogram.
294.
lines
the
.-.
OBA.
BC=AB.
AlsoZl
8.
XB
OBC=A
.-.A
7.
through
DO.
=
6.
and
=
C0
5.
A
1)0.
=
ODA,
2.
4.
In
tersect
in-
Only
BO
293.
BO
BD
O.
a
Argument
Ex.
and
diagonals AC
cannot
a
a
parallelogram is bisected,
parallelogram.
of two
angles of
bisect each
independent hypotheses which
quadrilateralis a parallelogram."
Construct
a
a
triangle
other.
would
lead to the
parallelogram,
givenits diagonals and
an
angle
92
GEOMETRY
PLANE
Proposition
242.
Two
included
parallelograms
angle of
the included
and
XXXVII.
one
are
Theorem
equal if two sides and the
equal respectivelyto two sides
are
angle of the
other.
II
To
II with
I and
ZU
Given
OI
prove
AB
=
EF,
AD
OII.
=
Argument
Place O
I
A
1. Transference
post. "54,14.
its
B
E,
upon
will
AD
linear with
Now
that
so
F.
upon
Point
II
fall upon
equal EF,
Then
Reasons
uponO
shall
AB
/.E.
EH, and Z.A=
=
IIAB,
DC
/.A
3.
AD
=
IIEF.
4.
By
def
DC
and
.:
DC
will become
HG
are
IIAB.
5.
AB
and
6.
Parallel
HG, and
somewhere
7.
By steps
H.
on
and
.-.
with
2.
ZE, by hyp.
=
EH.
will fall
D
col-
become
HG
both
collinear
.
of
^i*
by hyp.
a
" 220.
O.
coincide,
Arg. 1.
line
"179.
post.
will fall
C
HG
on
EH,
or
its
prolongation.
7.
Likewise
collinear
will
FG
8.
.*.
fall
its
or
will
BC
point C
point G.
with
become
FG, and
somewhere
and
C
.-.01
=
to
4, 5,
6.
on
prolongation.
must
fall
on
8.
Two
can
in
9.
similar
011.
Q.E.D.
9.
intersectingstr. lines
have
common.
only one
point
" 26.
By def. of equalfigures.
" 18.
CLASSIFIED
QUADRILATERALS
1.
93
I
BOOK
Opposite sides II: Parallelogram.
(a) Right-angled: Rectangle.
adj.sides equal:
Two
Square.
Khomboid.
(b) Oblique-angled:
adj.sides equal :
Two
Quadrilaterals
243.
Rhombus.
2.
other two
sides II,
Two
(a)
3. No
Ex.
show
is
a
If it is
298.
If
299.
only additional
sides equal.
the
are
three
non-ll sides
equal:
Isosceles trapezoid.
sides II: Trapezium.
Two
one
rightangle.
to be
proved
298
in Ex.
a
square,
is to prove
two
show
that
adjacent
a
rectangles and
Since
301.
Ex.
ezoid.
Trap-
given quadrilateralis to be proved a rhombus, what
steps corresponding to those given in Ex. 298 and Ex. 299 ?
If
300.
it has
step after those
the
Ex.
that
given quadrilateralis
a
:
required to prove a given quadrilaterala rectangle,
" 243 that the logicalsteps are to prove first that it
by reference to
parallelogram; then
Ex.
two
non-ll
rhomboids
parallelograms,they
are
ferentiate
property difgeneral propertiesof parallelograms. What
from
definition
?
rhomboids
rectangles
(2) by proof ?
(1) by
(See Ex. 266 and Ex. 267.)
all the
possess
Ex.
from
squares
(b) What
from
other
(c)
two
rhomboids?
Show
from
propertiesthat have been proved distinguish
other rectangles?
(See Ex. 277 and Ex. 278.)
two
properties that have been proved distinguishrhombuses
(a) What
302.
(See Ex. 277 and
propertieswhich
that the two
the other
Ex.
303.
from
Ex.
touch
304.
If
three
a
the sides of
of %B?
and
rhombuses
by
are
due
common
buses
rhomerty
prop-
a
distant
righttriangleis equi-
vertices.
given length is moved so that its ends always
given right angle, what is the locus of the mid-point
line AB
a
to the
and
definition.
mid-point of the hypotenuse of
The
the
278.)
distinguishsquares
of their class
members
possessed by squares
Ex.
of
94
PLANE
GEOMETRY
Proposition
three
If
244.
on
on
XXXVIII.
Theorem
parallel lines intercept equal segments
transversal, they intercept equal segments
or
one
more
other transversal.
any
IIlines
Given
AG,
CJ, DK,
BH,
etc., which
interceptthe
equal segments AB, BC, CD, etc.,on transversal AF, and
interceptsegments GH, HJ, JK, etc.,on transversal GL.
To
GH
prove
HJ
=
JK, etc.
=
Argument
Reasons
1. Draw
GM, HN, JR, etc. IIAF.
1.
Parallel line post.
2.
AG
2.
By
Now
MB, BHNC,
.'.GM
=
=
HN
AB,
BC,
=
The
JR
O
CD, etc.
4.
And
5.
.-.
6.
Again
AB
GM
are
CJRD,
BC
=
HN
=
def. of
a
Zl
.-.
GM,
=
Ax.
JR, etc.
HN,
JR,
8.
And
Z4
Z2
=
=
" 54, 1.
1.
If two
etc.,
str. lines
third
other.
Z5
Z3,
=
7.
etc.
Z6,
etc.
a
By hyp.
CD, etc.
=
IIto each
=
" 220.
O.
opposite sides of
are
equal. " 232.
str.
IIto each
7.
" 179.
HJ.
etc.,are
3.
which
are
other.
A have
IIright to
to
a.
" 180.
IIlines
their
sides
right and
left,they are
" 200,
a
line,they are
Corresponding A of
are
equal. " 190.
If two
IIto
left
equal.
I
BOOK
9.
A
.-.
GHM
A
=
AJKE,
HJN
9.
=
95
A
Two
and
etc.
.'.
GH=
HJ
10.
JK, etc.
=
Are
Question.
transversal
equal
the
to
the
adj.A
side and
of
the two
to
a
A
of the other.
Homol.
that the
segments
that
segments
one
adj.
" 105.
parts of
equal
figuresare equal. "
Q.E.D.
245.
the two
side
a
equal respectively
are
10.
equal if
are
110.
parallelsintercepton
they intercept on
another
one
versal
trans-
Illustrate.
?
246.
sides
of
bisects the
247.
bisecting one
of the non-parallel
parallel to the bases
trapezoid and
a
other
of
n.
The
Cor.
line
The
I.
Cor.
non-parallel sides
the
also.
line
joining the mid-points of
the non-parallel sides
of a
trapezoid is (a) parallel
to
the
bases; and
to
one
half their
Hint,
GH
Prove
equal
H
sum.
Prove
(a)
IICD.
(b)
WBC
EF
and
AD
then
AH=GB;
by
prove
the indirect method.
EF
=
GC
=
% (GC + HD)
KBC+AD).
248.
Cor.
side
one
parallel
the
249.
line
is
The
joining
the
.
of
Hint.
side
and
bisects
side
Draw
two
triangle
a
parallel
third
triangle
a
IV.
Cor.
of
line bisecting
side.
mid-points
sides
of
another
to
third
The
m.
to
and
CF
the
equal
IIBA.
to
one
Prove
half the third
CF=AE
=
EB.
(6) Draw
side.
=
96
PLANE
GEOMETRY
Proposition
To
250.
divide
XXXIX.
Problem
given straight line
a
into
number
any
of equal parts.
H
A
straightline
Given
To
divide
into
AB
B
AB.
equal parts.
n
I. Construction
1. Draw
the unlimited
2.
any
Take
lay it off
PH
n
convenient
times
n
3.
Connect
4.
Through
line AX.
segment,
AR,
as
and, beginning at A,
AX.'
on
the nth
as K, with B,
point of division,
as
precedingpoint of division,
P, draw
the
a
line
" 188.
IIKB.
5.
Then
6.
.-.
is
HB
nth
one
HB, if laid off
of
AB.
on
successively
AB, will divide
AB
into
equal parts.
II. The
proof and discussion
left
are
as
an
exercise for the
student.
Ex.
305.
Divide
Ex.
306.
Construct
an
Ex.
307.
Construct
a
The
line
251.
Def.
parallelsides
Ex.
308.
be obtained
of
a
a
straightline
into
7
equilateraltriangle,having given
square,
having given
Cor.
the
I -and
Cor. IV
from
Cor.
the
the
the
Ill,Prop. XXXVIII,
Cor. II.
eter.
perim-
perimeter.
joining the mid-points of
of
trapezoidis called the median
that
Show, by generalizing,
from
equal parts.
non-
zoid.
trape-
may
98
GEOMETRY
PLANE
Proposition
The
252.
from
and
Z ABC',
PD
prove
P
any
A
the bisector of
and
respectively.
BC
sides
of
Ex.
from
the
3.
.-.
ADBP
4.
.'.
PD
XL
the
Ex.
other
also at
and
Ex.
and
two
sides
given
Find
of
follows
as
an
:
angle is equidistantfrom
a
a
the
trianglewhich
from
two
Construct
the
always
equidistant
a
given intersectinglines
two
line.
fixed third
lines
given intersecting
two
given parallellines.
the
altitudes of a'rhombus
327.
is
triangle.
point equidistantfrom
two
that it
side of
one
point equidistantfrom
a
Ex.
prove,
PB.
Q.E.D.
of the
distance
The
so
APBE.
point equidistantfrom
a
326.
rolls
=
be stated
point in
a
Ex.
which
PBE.
=
=PE.
equidistantfrom
325.
Z
PB
PBE,
angle.
Find
324.
also
Ex.
a
"
may
Find
323.
PD
Only
and
DBP
in the bisector
Find
322.
rt. A
In
DBP
Every point
ZABC;
PE.
2. Z
Prop.
angle
an
equal.
are
Argument
253.
of
point in BR,
to BA
P
=
1.
sides
D
PE, the is from
To
to the
its bisector
in
B
Given
Theorem
perpendiculars
two
point
any
XL.
locus of the center
touches
the
four sides of
are
of
sides of
equal.
a
rhombus.
Prove.
a
circle of
a
givenangle.
given radius,
Do
not
I
BOOK
99
XLI.
Proposition
Theorem
(Oppositeof Prop. XL)
The
254.
from
point
any
in
not
the
to
perpendiculars
two
its bisector
are
of an
unequal.
sides
angle
C
/.ABC;
Given
and
PD
To
and
to BA
P
Outline
Draw
Now
J_
FG
255.
Prop.
the sides
256.
Cor.
257.
Then
PG.
XLI
in
not
of
be
may
FE
=
FG.
But
the
stated
the bisector
PG
"
PD.
follows
as
of
an
angle is
(Converseof Prop. XL).
the
:
not
equidistant
angle.
sides
of
an
Prove
angle lies
equiEvery point distant
in the bisector of
Cor.
directly,using the figurefor " 262, or apply " 137.
II.
The
bisector
points equidistant from
Ex.
a
Proof
of
angle.
Hint.
all
respectively.
BC
.'.PE"PG.
PG.
I.
from
the
/ABC;
PD.
"
Every point
from
; draw
BA
PF-\-FG"
.'. PE
in BR, the bisector of
pointnot
any
A
=" PE.
PD
prove
P
J" from
PE,
D
G
B
328.
What
is the
lines ?
pair of intersecting
angle is the locus
the sides of the angle.
of
locus of all
an
points
that
are
of
equidistantfrom
100
PLANE
GEOMETRY
.CONCURRENT
LINE
Proposition
bisectors
The
258.
XLII.
C
is
a
A
Given
AB
with
O
Theorem
of the angles of
point which
sides of the triangle.
in
THEOREMS
AF, BE,
CD
current
triangle are conequidistant from the three
the
a
respectively.
To
(a) AF,
prove:
BE,
CD
in
concurrent
Argument
and
BE
pointas
some
at
1.
0.
If two
AB, BC, and
a
str. lines
transversal
the
2
from
0
to
OG, J"
AB, BC, and
2.
V
0
is in BE,
OL
one
=
OH.
a
3.
The
an
its
int. A
on
of
the
equal
not
lines
the
point outside
there exists
CA
respectively.
3.
From
and
one
J_ to the line.
two
Z
the
to
are
" 194.
II.
not
OL, OH, and
A,
rt.
by
making
side
same
transversal
Draw
CA.
cut
are
of the two
sum
2.
0;
Reasons
will intersect
CD
point as
some
(b)the point 0 equidistantfrom
1.
A, B, and
of A
bisectors
Js to the
from
any
bisector
are
a
line
only
" 155.
sides
of
point in
equal.
"252.
0
OL
is
in
=
CD, OG=OH.
OG;
4.
Same
5.
Things equal to the
thing are equal to
other.
" 54; 1.
reason
as
3.
same
each
aoi
i
book
Reasons
Argument
6.
.*
AF, the bisector of Z.CAB,
.
through
passes
6.
the
lies in
the Z.
7.
AF, BE, and
.-.
CD
are
0
is
8.
equidistantfrom
AB, BC, and
of
Z
an
bisector
of
" 256.
lines.
of concurrent
.
"196.
in 0.
8. Also
def
By
7.
current
con-
sides
the
from
O.
equidistant
point
Every
CA.
By proof,OL
OH=
=
OG.
q.e.d.
point of intersection of the bisectors of
the three angles of a triangle is the locus of all points
equidistant from the three sides of the triangle.
259.
Is it
329.
Ex.
The
Cor.
given straightlines
sides of
Ex.
a
point
a
EBC,
three
lines
angle
BAC
and
sides AB
AE, AF, and BC.
equally distant
and
from
a
are
triangleABC
anglesBAC,
bisectors of the three
other
Is any
of
AG
in
point
the
lines?
three
these
equallydistant
is
point which
a
?
perpendiculars from it to three
(Give geometric construction.)
equal.
through
three
the
that
the
F, respectively,
all pass
BCF
be
point equidistantfrom
a
given straightlines
such
that if the
Prove
to E
and
your
four
quadrilateralshall
331.
prolonged
? from
Find
330.
Ex.
find
always possibleto
from
bisector
Give
the
of the
for
reason
answer.
Through a given point P draw
perpendiculars to it from two fixed points Q
332.
Ex.
equal
segments from
which
rolls
so
that
it
See
(Hint.
Construct
333.
Ex.
P.
always
shall cut
B
such
off
that
it
on
" 246.)
the locus of the center
of
the sides of
touches
straight line
a
and
a
circle of
givenradius,
given triangle. Do not
a
prove.
Ex.
Find
334.
the locus of
a
point in
equidistantfrom two other sides.
of the parallelogram?
vertex
Ex.
Find
335.
equidistant from
class of
Ex.
which
prove.
the
locus of
of
two
parallelogramsis
Construct
336.
rolls
so
that it
the
a
In
what
point in
vertices
this locus
a
of
vertex
the locus of the
constantly touches
side of
one
parallelogramand
this locus
parallelograms is
side of
one
the
of
center
a
a
a
parallelogramand
parallelogram. In
the parallelogram?
of
given
a
a
circle of
what
given radius
circumference.
Do
not
GEOMETRY
PLANE
Proposition
The
260.
the three vertices
A
Given
To
in
prove
equidistantfrom
FG, HK, ED, the _L bisectors
(a) FG,
:
is
point which
a
ED
HK,
in
concurrent
and
will intersect at
ED
pointas
some
1.
Draw
3.
v
0
OC.
is in FG, the _L bisector
v
OB
AB,
0
.-.
OB
5.
.-.
HK,
=
is in DE,
BC, passes through
6.
.*.
FG, HK, and
ED
are
3.
Every point
of
of
4
Ax.
5.
Every
6.
0
is
C.
" 54, 15.
the
sector
_L bi-
line is
a
distant
equi-
the ends
point equidistant
the
ends
of
a
line
lies in the
_L bisector
that line.
" 139.
def
q.e.d.
7o
of
" 134.
By proof,OA=OB
of
lines
of concurrent
.
"196.
equidistantfrom
A, B, and
By
" 195.
" 54, 1.
1.
from
0.
current
con-
in
from
that line.
in 0.
J. Also
post. I.
Str. line
OA.
_L bisector
to
respectively
2.
sector
the _L bi-
OC.
the
C.
intersect.
and
OA;
=
of CA, 0C=
4.
0 ;
intersecting lines
two
0.
OA, OB, and
of
A, B, and
lines X
Two
also
2.
pointas
Reasons
Argument
FG
of AB, BC, CA.
some
(")the point 0 equidistantfrom
1.
angle
tri-
a
of the triangle.
with
ABC
Theorem
bisectors of the sides of
perpendicular
concurrent
are
XLIII.
=
OC.
BOOK
103
I
point of intersection of the perpendicular
bisectors of the three sides of a triangle is the locus
of all points equidistant from the three vertices of the
triangle.
261.
Ex.
The
Cor.
Is it
337.
given points ? from
triangle,and
acute
four
given points
then
construct
through
the
vertices of the
Ex.
339.
Construct
the
of
vertices
a
the vertices of
a
The
Given
A
To
prove
lines
of
circle whose
a
obtuse
an
ABC
with
circumference
an
shall pass
shall pass
through
circumference
shall pass
through
the
the sides of the
concurrent.
AD, BE, and
of
A, B, and
Proof
C, of
triangleABC,
respectively. Then
AB,
CF
auxiliaryA
concurrent.
CF,
concurrent.
CF
vertices
Theorem
triangle are
a
its altitudes
AD, BE, and
are
sides of
circumference
XLIV.
of
/I7,that AD, BE, and
CF
two
triangle.
altitudes
||BC, AC, and
BE^sad
circle whose
of
righttriangle.
Outline
Through
three
?
circle whose
Proposition
262.
point equidistantfrom
triangle.
a
Construct
340.
Ex.
a
perpendicular bisectors
the
Construct
338.
Ex.
always possibleto find
are
the _L
HKL.
draw
by means
of
bisectors,respectively,
prove,
Then, by Prop. XLIII, AD,
q.e.d.
104
GEOMETRY
PLANE
Proposition
Any
263.
in
other
A
Given
To
OD
trisection
that
prove
and
\AD
=
with
ABC
and
AD
OE
and
AD
i
=
CE
intersect
CE
and
2.
Let
3.
Quadrilateral REDS
4.
.-.
5.
That
CE
RO=
AR=
is, OD
"
=
Cor.
T/^e
265.
Def.
The
called
of
or
mass
Ex.
its
341.
sides,and
Ex.
the
342.
center
Draw
repeat
Ex.
343.
drawing
Ex.
angle
Prove
lines to B
344.
between
of
and
AO
SO
=
0^
=
center
of
a
the
of
the
a
them.
a
triangle
of the
the
end
of the
triangle,given
of
a
triangle. It is also
the
center
will intersect
on
will
outside
by prolonging OD
XLV
current.
con-
are
medians
point is
triangle.
altitudes
trianglewhose
the proof for Prop. XLIV.
Construct
respectively.
triangle. This
gravityof
C from
CO
q.e.d.
intersection
median
Prop.
" 194.
0.
J-CB.
a
and
that
OE.
=
altitudes
triangle whose
repeat the proof for Prop. XLIV.
Draw
triangle,and
point as
some
medians
of the
of
CS
and
AD
point
the centroid
such
point O
O.
a
and
""ree
called the
at
mid-pointsof
is
OD
264.
triangleis
the
S be
and
R
a
Proof
of
will intersect
AD
in
CE.
Outline
1.
of its medians.
two
any
each
intersect
of a triangle
point of each.
medians
two
a
Theorem
XLV.
intersect
its
own
one
length
of
of
and
prolongation.
two
of
its medians
and
the
106
PLANE
GEOMETRY
CONSTRUCTION
OF
TRIANGLES
A
269.
triangleis determined,in general,when three parts are
given,provided that at least one of the given parts is ar line.
The three sides and the three angles of a triangleare called
its parts ; but there
medians, the
into which
three
both
also many
parts ; as the three
altitudes,the three bisectors,and the parts
are
the
sides
and
indirect
the
divided
angles are
by
these
lines.
A
A
The
270.
used
for
given
in the
annexed
figuresmay
be
brevity:
angles of the triangle; in
is the rightangle.
the
A, B, C,
C
a,
notation
the sides of the
b, c,
triangle;in
a
a
righttriangle,
angle
c is
righttriangle,
the
hypotenuse.
mai mb, mc, the
medians
to
a,
b,and
ha,hbihc,
the altitudes
to a,
ta,tb,tc,
the bisectors
of A, B, and
lu,da,
the segments
sa, ra,
the
The
271.
of
segments of
student
should
c
b, arid
respectively.
respectively.
c
respectively.
C
a
made
by
the altitude to
a,
a
made
by
the bisector
angle A.
review
the chief
trianglesalready given : viz.
:
right triangles
a, b ; b, c ; b, B ; b,A ;
of
Ex.
347.
State
in words
a, B,
cases
c
;
given
of construction
A, b, C ; a,b,c:
c, A ; review
the first eight cases
of
also " 152.
in " 271.
BOOK
To
construct
Problem
XLVII.
Proposition
272.
107
I
triangle having
a
of its sides
two
the angle
to two given lines,and
equal respectively
one
of these lines equal to a given angle.
site
oppo-
Y
B
1.
Fig.
lines
Given
To
b, and Z.B.
and
a
A
construct
ABC.
I. Construction
A
1.
Draw
2.
At
line,as BX.
point in BX, as B,
any
any
3.
On
4.
With
.BFlay off
C
cuttingBX
5.
A
II.
BC
=
to the
given
and
center
as
a.
=
with
is the
ABC
proof is
The
(1) If
b "
be
can
be the
radius,describe
as
an
arc
requiredA.
left
as
b
same
=
Discussion
greater than a;
may
be
(2) b
(3) b
may
equal a
may
be less than
a, there
will be
constructed
in
a, the A
as
exercise for the student.
an
(1) b
is shown
case
(2) If
b
at A.
III.
This
XBY
" 125.
B.
one
Z
construct
for
case
which
one
;
a.
i.e.
solution,
shall
contain
one
the
A
and
only
given parts.
Fig. 1.
will be
(1).
isosceles.
The
construction
will
108
GEOMETRY
PLANE
A
B
(3) If
b "
the _L from
and
Cto
If b "
rt. A.
the
In
the _L from
thus
discussion
The
in which
273.
angle
contain
it is
is either
required parts.
solution.
one
BX, there
will be
the
case
Construct
348.
a,
ha, ma.
Imagine the problem solved as in
Fig. 1, and mark the given parts with heavy lines.
be made
The
is determined
and may
triangleAHM
Analysis.
the
basis of the
Ex.
Analysis.
AOC
construction.
Construct
349.
From
be
may
known, since AO
Ex.
Analysis.
In
determined.
CB,
it and
is
CB
and
a
CO
that
seen
sides
Its three
"
f
mc.
triangle,
given a, ha,hc.
Fig. 3, right triangleCHB
The
parallelto
it will be
constructed.
\ma
"
triangle,given 6, ma,
Fig. 2
Construct
350.
a
so
locus
of
that
the
equal to hw
ABC
equals
will be
vertex
a
solution.
acute.
was
is
a
rt.
the
Z
given
?
triangle,given
a
A
possiblewhen
followingexercises are given to
illustrate analysisof problems and to show the
of auxiliary
use
trianglesin constructions.
Ex.
no
given Z
given Z
The
274.
The
If b
b "
is left to the student.
(1) the only
is
obtuse
right or
Z
(1). If
case
solutions,since A
two
in which
cases
Why
for
as
considered,the
obtuse
an
Question.
C to
far
of the
be
the
there will be
BX
cases
will
there
BX,
Fig. 2, both
BBC,
the _L from
and
to
C
the construction
Make
a.
A
distance
is
a
is
line
between
mc
angle
triare
j
I
is
and
Construct
the
angle between
and
ma
~?B
\
a, "na,
^"'
/
*"
a.
K
determined,
G is
as
Fig.
4.
Fig. 5.
in
shown
CB.
bisects
triangle,given
a
Triangle AM
Analysis.
Since
2raa.
parallelogram, AK
a
352..
Ex.
c, and
sides, b,
three
by
,c, ma.
mined
Fig. 4, is deter-
Triangle ABK,
Analysis.
ABKC
atriangle,given b
Construct
351.
Ex.
109
I
BOOK
A
M
rr~v
"
S
Fig.
Construct
353.
Ex.
Construct
354.
Ex.
the perimeter,
between
B
construct
be
may
them
angle,and
The
determined.
is
parallelogram,given
a
base
one
Construction.
AJE, Fig. 7,
Fig. 6,
Triangle EST,
Analysis.
drawn
that the distance
so
equals the altitude
equal
angle EBC
; at
any
point
to
the
given
jj
2"37
and
CH
parallels,
two
q
the altitude.
B-
A
E
FlG-
""
M
M
R
K
CA, bisectingangle FOB
angle ; draw
;
half
the
AE
to
equal
given perimeter ;
base
measure
f
complete parallelogram CHEB.
Ex.
two
N
6.
sides.
trapezoid,given its four
a
d
Construct
355.
non-parallelsides
Construct
the
difference between
N
Fig.
trapezoid,given the
a
and
S
the bases.
8.
(See Fig.8.)
triangle,given :
a
Ex.
356.
A, ha, ta.
Ex.
360.
B, ha b.
Ex.
357.
ha} la,da.
Ex.
361.
a", ta, ha.
Ex.
358.
ha, ma,
Ex.
362.
a, mb,
Ex.
359.
a, U,
Ex.
363.
a, U, B.
Construct
la.
C.
isosceles
an
Ex.
364.
The
two
Ex.
365.
One
base,
Ex.
366.
A
C.
trapezoid,given :
bases
the
and
the
altitude.
altitude,and
base, a diagonal, and
the
a
diagonal.
angle between
them.
110
PLANE
FOR
DIRECTIONS
I.
275.
Make
GEOMETRY
THE
the
SOLUTION
EXERCISES
OF
figuresclear,neat, accurate,
and
general
in character.
Fix
II.
firmlyin
to the
in
fundamental
in
to show
try
erence
ref-
to
the proposition
you,
contradict
\
find
can
the conclusion
and
propositionsrelated
question.
If you
IV.
with
conclusion
hypothesisand
given figure.
Recall
III.
mind
every
the
no
theorem
which
possible way
helps
{reductioad
absurdum)
the contradiction.
absurdityof
Make
V.
sists
frequent use of the method of analysis,which conassuming the propositionproved, seeing what results
in
until
follow
a
known
truth
is
retracingthe
reached, and
then
find
which
steps taken.
VI.
it is
If
conditions,it
required
is often
to
point
a
convenient
point by the
Intersection
of Loci.
By finding the locus of a point which
satisfies each condition
separately,it is possibleto find the
points in which the two loci intersect;i.e. the points which
satisfyboth
See
VII.
conditions
at the
" 152
exercises
and
attackingproblems
The
method
uncertain
issue
to
time.
same
just
described
to
certain
a
following "
under
is
V
/Substitutions.
is true
if .B.istrue.
3.
But
2.
B
is true
if
4.
.-.
form.
is true.
is also called
proofs of
the theorems
But
these
for method
of
were
rearranged in the form
the
are
a
shiftingof
Analytic
A
C
the
is true.
of
is called
proof. The
the Synthetic
first thought
then
through analytically,
in which
find them.
we
:
is true.
Method
put in what
called
an
be illustrated thus
It may
A
This
274
It is sometimes
one.
1.
C
the
of construction.
of Successive
Method
find
fulfills two
BOOK
MISCELLANEOUS
of
side
Ex.
368.
Ex.
369.
Construct
Ex.
drawn
from
angle of 75"
an
side
that
upon
; of
the
Construct
371.
Construct
one
equal.
are
97^".
perpendicufars
that
a
370.
of
extremities
given line find a point such
sides of an angle shall be equal.
Upon
it to the
from
median
the
triangleto
a
EXERCISES
perpendiculars
The
367.
Ex.
111
I
triangle,given its perimeter and
two
of
parallelogram,
given
base
angle,
a
its
angles.
Ex.
and
a
bisector of the base
the
Given
372.
Ex.
Construct
Ex.
of their
the bisector
Construct
373.
side,and
the
if
meet
of the
other
sides.
two
given angle less than the other unknown
opposite the given angle greater than the other
The
374.
difference
is 90" ; find all the
Ex.
A
375.
Represent
the
answer
Ex.
a
no
376.
Ex.
two
adjacent
side
2
site
oppo-
side
The
:
side.
unknown
a
gram
parallelo-
and
2 miles
passes
and
town
find
by
from
certain
a
1 mile
from
construction
how
A
town.
the
railway.
places
many
through
lie outside
given points which
two
line.
Show
when
379.
380.
sides of
makes
Two
and
the
sides.
two
If two
right triangle,given the hypotenuse
a
a
triangleare
the greater
trapezoidsare
angle
unequal, the
with
equal if
the
their
median
through
lesser side.
sides taken
in order
are
to each.
Construct
a
righttriangle,having given
its
perimeter and
angle.
381.
Draw
a
line such
given indefinite lines shall
Ex.
Case
of the circle to be in that
center
of the other
equal, each
acute
side.
The
adjacent angles of
two
the
circle
a
Construct
intersection
Ex.
:
is possible.
378.
Ex.
from
point
a
Describe
377.
Ex.
Ex.
by
given line,the
difference
an
as
1
description.
solution
their
4 miles
town
angle, one
angles.
straightrailway
place is described
the
between
one
Case
the
Ex.
sufficiently
prolonged.
prolonging the lines.
angle, without
triangle,having given
a
difference
base, one
angle.
lines that would
two
the
382.
One
angle of
be
that
its segment
intercepted between
equal and parallelto a given finite line.
parallelogramis given in position and the
the parallelopoint of intersection of the diagonals is given ; construct
gram.
a
112
PLANE
Construct
383.
Ex.
GEOMETRY
the median
sides and
given two
triangle,
a
to the
third side.
If from
384.
Ex.
of the
vertices
three
point within
the sum
of
triangle,
the, sides of the triangle,and
of
vertex
of
isosceles
at
an
each
times
with
of
end
one
through
the
Ex.
387.
in D, and
388.
the
lines drawn
the
point.
Ex.
the
base
Fig.
triangle,and
Fig.
1.
of the
the
prolongation
equilateraltriangle.
an
of the
bisector
The
DE
sum.
at
forms
base
vertex,
of the exterior
Ex.
of
of
angle,
the
to
the
side
one
sum
proof
base
perpendicular
the
the
angle at the
triangle
If the
is four
half their
to
2.
386.
Ex.
drawn
are
lines is less than
these
greater than
cases
and
using Figs. 1
once,
the
for two
XIX
Prop.
the
Repeat
385.
Ex.
trianglelines
a
any
is drawn
parallelto
angle
G in F.
at
angle
AG
a
locus.
from
a
given point
389.
Construct
an
390.
Construct
a
391.
The
of
Find
DE
the
to
a
=
triangleABC
a
meeting
Prove
Define
G
BO
other
side
meets
AB
the bisector
and
in E
2.
EF.
mid-points of all
not passing through
locus
of the
given
line
and
one
diagonal and
one
isosceles trapezoid,given the
bases
angle.
Ex.
square,
given the
of
sum
a
side,
Ex.
prolonged
of
isosceles
an
of the distances
difference
triangleto
the
sides
point in the base
of the triangle is
two
lines
intersecting
from
equal
any
constant.
Ex.
and
at
Ex.
two
Find
392.
a
a
point
from
given distance
When
393.
two
equidistant from
X
a
given point.
lines
are
met
corresponding angles is equal to
Construct
a
the
by a transversal,the difference of
the two lines. %
angle between
triangle,given ",
Ex.
394.
A, ha, la.
Ex.
398.
a,
Ex.
395.
A, ta, sa.
Ex.
399.
mc,
Ex.
396.
a,
Ex.
400.
6, c,
Ex.
397.
a, 6 + c, A
Ex.
401.
A, B,
ha, la.
m.
hCJ B.
b
-TcT
114
PLANE
GEOMETRY
I.
Proposition
diameter
Every
284.
of
Theorem
circle
a
bisects the
ence
circumfer-
the circle.
and
M
circle
Given
To
prove:
with
AMBN
O, and
center
diameter.
AB, any
(a) that
AB
bisects circumference
(b)that
AB
bisects circle AMBN.
AMBN)
Argument
figureAMB
Turn
Reasons
AB
on
as
it falls upon the plane of
Arc AMB
will coincide with
.-.
AMB
arc
arc
=
circumference
4.
axis until
an
" 54, 14.
2.
" 279, a.
3.
" 18.
4.
" 279,
5.
" 18.
ANB.
ANB.
arc
bisects
; i.e. AB
ANB
1.
AMBN.
will coincide
Also figureAMB
with
figure
a.
ANB.
5.
.'.figureAMB
bisects circle
Ex.
402.
rectangleas
Ex.
403.
A
figure ANB',
=
AMBN.
diameters.
into four
Construct
Ex.
405.
Construct
given
the
Prove
semicircles
a
points A
circle
and
B
of the
diagonals of
a
equal.
each
other
divide
a
ference
circum-
by superposition.
circle which
a
each
upon
perpendicularto
Prove
equal arcs.
404.
through
is described
diameters
Ex.
two
Q.E.D.
semicircle
Two
i.e. AB
shall pass
having
a
through two given points.
given
radius
r, and
passing
285.
is
286.
to, or
touches,
287.
sector
as
as
290.
a
Ad
major
semicircumference,
a
greater than
arc
arc
as
a
DME\
minor
central
is at
vertex
circle,
semi-
a
is called
arc,
A
Def.
is
is half
which
arc
is called
whose
ties,
extremi-
AMB.
An
Def.
a
circle
a
circle is called
Def.
291.
of
segment which
A
circumference
292.
radii
plane closed figurewhose
and their interceptedarc,
a
DCE.
segment
as
called
circle is
a
of two
joiningits
Def.
half of
a
of
sector
segment
segment
289.
one
at
figure whose
composed of an arc
the chord
and
0
closed
is
boundary
of
A
Def.
plane
a
to circle
point of tangency.
composed
is
tangent
SOB.
288.
is
A
is
HK
tangency.
is the
T
Def.
boundary
of
point
point T, and
point is
This
point.
the
called
in
circumference
the
one
cle
cir-
a
prolonged,
far
if,however
it meets
MN.
straight line
A
Def.
is tangent
but
terminated
is not
circumference, as
the
cle
cir-
a
in two
circumference
points,but
by
of
secant
straight line which
a
the
cuts
A
Def.
115
II
BOOK
arc,
as
of
the center
a
less than
arc
is
cumference
semicir-
a
DCE.
arc
angle
angle, or
AMB.
arc
semicircumference
a
an
as
at
the
center, is
circle and
whose
an
angle
sides
are
radii.
Ex.
8 and
406.
The
line
joiningthe
10,respectively. What
..Ex. 407.
A
circle
can
are
have
centers
the
only
of two
relative
one
circles is 6, the radii
positionsof
center.
the
two
are
circles ?
"
116
PLANE
GEOMETRY
Proposition
II.
Theorem
circle,if two central
equal circles,or in the same
cumferenc
angles are equal, they interceptequal arcs on the circonversely,if two arcs are equal, the central
angles that intercept them are equal.
In
293.
I.
and
To
equal circles ABM
Q, interceptingarcs AB and
"b
CD.
prove
Given
and
CDN, and
CD,
equal central
=
Place
circle
ABM
that center
and
2.
A
3.
OB
4.
.-.
5.
.-.
6.
.'.
QA
O
circle CDN
upon
shall fall upon
center
collinear with
fall upon
will coincide
AB
AB
=
with
so
CD.
"54,14.
2.
"
3.
By hyp.
4.
"
QC.
QD.
D.
with
1.
Q,
C.
will
B
Reasons
shall be collinear
will fall upon
will become
II.
CD.
Q.E.D.
279, b.
5.
279, 6.
" 279, b.
6.
"18.
Conversely:
and CDN, and equal arcs
equal circles ABM
intercepted
by A 0 and Q, respectively.
Given
To
G
respectively.
Argument
1.
A
prove
Zo
=
ZQ.
AB
and
CD,
BOOK
117
II
Argument
circle
Place
ABM
that center
2. Kotate
until
OA
ABM
C, B upon D.
upon
will coincide with
Z
.-.
0
294.
Z
=
its
QCand
CDN
In
so
center
as
equal
QD.
Q.E.D.
equal circles, or
"54,14.
2.
"54,14.
3.
"17.
4.
"18.
A
CD,
with
OJ?
1.
Q.
pivot
a
Q.
Cor.
central
O
upon
falls upon
AB
circle
upon
shall fall upon
0
circle
Reasons
in
the
sams
circle, if
cepts
unequal, the greater angle interthe greater arc; conversely,if two arcs
are
unequal,
central
angle that intercepts the greater arc is the
two
the
angles
greater. (Hint.
295.
Lay off
fourth
A
Def.
are
the
smaller
part of
central
a
angle upon
circumference
the
is
greater.)
called
a
quadrant.
Prop. II it is evident
that a right angle at the center
cumference.
interceptsa quadrant on the cireters
Thus, two " diamFrom
and
AB
circumference
into four
AC, CB, BD, and
296.
Def.
A
by
a
central
the
quadrants,
DA.
degree
degree, is the
arc
divide
CD
arc
angle of
of arc, or
an
intercepted
one
degree.
A
right angle contains ninety angle degrees (" 71);
therefore,since equal central anglesinterceptequal arcs on the
circumference,a quadrant contains ninetyarc degrees.
Again, four rightangles contain 360 angle degrees,and four
cumferen
right angles at the center of a circle intercepta complete circontains 360 arc degrees.
; therefore,a circumference
contains 180 arc degrees.
Hence, a semicircumference
297.
Ex.
408.
equaJ .arcs.
Divide
a
given circumference
into
eightequal arcs
; sixteen
118
PLANE
Ex.
equal
Divide
409.
arcs
Ex.
twelve
;
410.
equal
into two
Ex.
Ex.
a
and
B
If
412.
by
Ex.
a
a
diameter
If
413.
a
the
its center
In
chords
are
if two
arcs
shall pass
given
a
another
and
diameter
to
through
line
arcs
three
;
it divide
a
cumference
cir-
chord
another
which
chord
two
given points
c.
are
interceptedby the angle
drawn
parallelto the chord.
are
bisects their
drawn
from
between
drawn
a
point
in
will
be
them
from
point in
a
will be
interceptedarc
chord.
Proposition
298.
equal
arcs.
in
and
diameter
the
circumference,
parallelto
arc
six
perpendicular
secant
a
circle which
diameter
a
circumference, the
bisected
and
a
shall have
into
arcs.
pairs of equal
Construct
411.
and
A
given circumference
a
diameter
A
GEOMETRY
III.
Theorem
circle, if two
equal circles,or in the same
equal, they subtend
equal arcs;
conversely,
them
are
are
equal, the chords that subtend
equal.
I. Given
equal circles
To
AB
prove
0
and
Q, with
equal chords
Reasons
radii OA, OB, QC, QD.
Draw
2.
In
A
3.
OA
=
4.
.-.A
5.
.-.
Z0
6.
.'.
AB=
and
OAB
QC
OAB
=
and
=
Z
CD.
QCD,
OB
A
and
CD.
=
Argument
1.
AB
=
AB
=
CD.
QD.
QCD.
Q.
Q.E.D.
1.
" 54, 15.
2.
By hyp.
3.
"279,
4.
"116.
5.
"110.
6.
" 293, I.
b.
CD
Conversely:
II.
equal circles
Given
To
119
II
BOOK
chord
prove
AB
Q, and
and
0
chord
=
equal arcs
and
AB
CD.
Argument
Draw
2.
AB
3.
.*.
4.
OA
5.
.:
A
6.
.-.
chord
the
Zboa
By hyp.
Zdqc.
=
and
QC
=
OAB
OB
A
=
^4i?
If
414.
"107.
chord
CD.
415.
A
Ex.
416,
If two
equal, its diagonals are
Def.
and
the
prove
number
into any
in
a
be
an
a
equal arcs,
rectangle.
inscribed
of Ex.
converse
of
equal.
circle is
opposite sides of
equal.
quadrilateral
416.
is inscribed
A polygon
if all its vertices
circle
a
the
State
417.
299.
in
is divided
parallelogram inscribed
of the
"110.
Q.E.D.
circumference
Ex.
Ex.
" 279, b.
QD.
=
QG'D.
=
a
" 293, II.
joiningthe points of division will
chords
are
" 54, 15.
QC, QD.
CD.
=
Ex.
Keasons
radii OA, OB,
1.
CD.
are
on
circumference.
is
ABODE
300.
an
Def.
in
a
Thus, polygon
inscribed polygon.
If
is
polygon
a
circle,the
to be circumscribed
circle
about
scribed
in-
is said
the
gon.
poly_^
Ex.
418.
Inscribe
an
equilateralhexagon
in
a
circle ;
an
equilateral
triangle.
Ex.
419.
The
Ex.
420.
If the
joined,
will the
Ex.
an
diagonalsof
inscribed
State
StafceVthetheorems
extremities
inscribed
of any
rectangle will be
rectanglebe
421.
an
a
square
may
are
equal.
intersectingdiameters
two
formed.
what
Under
are
conditions
?
the theorems
which
equilateral
pentagon
which
be
used
may
in
be used
in
proving arcs
proving chords
equal.
equal.
120
PLANE
GEOMETRY
Proposition
are
unequal,
arc
;
the
greater chord
conversely,if
the
that subtends
greater
I.
Given
circle 0, with
To
prove
AB
circle,if two
same
the
subtends
minor
two
Theorem
in the
equal circles,or
In
301.
IV.
arcs
greater minor
unequal, the chord
are
is tliegreater.
arc
chord
AB
"
chord
CD.
CD.
"
Argument
Draw
2.
In
A
OAB
3.
Chord
AB
4.
.".
Zl
"
Z2.
5.
.*.
AB
"
CD.
II.
and
OA
OCD,
chord
"
OD
=
CD.
Q.E.D.
1.
"
54, 15.
2.
" 279,
3.
By hyp.
4.
" 173.
5.
" 294.
circle 0, with
chord
prove
AB
AB
CD.
"
chord
"
CD.
Reasons
Argument
1.
Draw
2.
In
3.
AB
4.
.\^C"
5.
.-.
Ex.
a.
Conversely:
Given
To
Reasons
radii OA, OB, OC,
1.
chords
radii OA, OB, OC,
A
and
OAB
OCD,
1. "
OD.
OA
=
OC,
OB
OD.
"
"/SU^
chord
422.
Z2.
AB
Prove
"
chord
the
converse
CD.
q.e.d.
of Prop. IV
by
the
54, 15.
2.
" 279,
3.
By hyp.
4.
"294.
5.
" 172.
indirect
a.
method
"
122
GEOMETRY
PLANE
Ex.
424.
Through
a
shall be bisected
which
Given
425.
Ex.
426.
Ex.
; find
Ex.
chords
Ex.
A
diameter.
2.
A
perpendicular to
of
chord.
A
A
bisector
of the
major
arc
of
a
chord.
5.
A
bisector
of the
minor
arc
of
a
chord.
of
circle
chords
line
passes
through
428.
Given
a
passing through
the
an
a
of the
center
arc
of
a
the
in
positionand magnitude
VI.
given
middle
points
of two
parallel
circle.
circle,find
a
Proposition
bisect
given
are
circle.
of the
center
tions,
following condi-
chord.
3.
two
chord
a
three:
4.
The
To
five
bisector
427.
306.
a
circle construct
a
of the
two
remaining
1.
Any
the
fulfilling
any
line
a
it fulfillsthe
that
prove
at
within
given point
the point.
the center
of the circle.
Problem
arc.
A
Given
AB,
bisect
To
The
using a
the
arcs
"
circle.
of any
AB.
and
discussion
left
are
as
an
for the student.
Construct
429.
Ex.
Ex.
arc
construction,proof,
exercise
30"
an
twice
radius
as
-an
arc
long
of 45" ; of 30".
as
the
one
Construct
previously used.
an
arc
Are
of
30",
these
two
"
and
equal ?
430.
center
Distinguishbetween
of
an
arc."
findingthe
"
mid-point
of
an
arc
BOOK
Proposition
123
II
VII.
Theorem
circle, if two chords
equal circles,or in the same
versely,
are
equal, they are equally distant from the center; conif two chords are equally distant from the center,
they are equal.
307.
In
circle
I. Given
OF
of
be the distances
To
OE
prove
with
O
chord
and
AB
AB
chord
=
from
CD
CD, and
center
2.
Reasons
and
.*.
in rt. A
OB
F
.'.
A
.'.
OE"
=
CF.
A
=
OCF.
OF.
Q.E.D.
circle
Prove
431.
432.
0
with
OE, the distance
the distance
chord
prove
inscribed
Ex.
EB
OCF,
and
AB
1.
" 54, 15.
2.
" 302.
3.
" 54, 8
4.
" 279,
5.
" 211.
6.
" 110.
a.
a.
Conversely:
Hint.
an
and
OEB
equal to OF,
Ex.
OC.
the
are
OEB
Given
To
and
OC.
=
II.
0,
OB
mid-pointsof
CD, respectively.
E
and
OF.
=
radii
OE
respectively.
O,
Argument
1. Draw
let
A
AB
OEB
of chord
chord
=
A
=
CD
of chord
AB
from
center
from
0.
CZ).
OCJl
If
perpendicularsfrom the center of a circle to
polygon are equal, the polygon is equilateral.
If
through
making equal angles with
any
the
center
point in a diameter
diameter, the two
two
chords
chords
are
the sides of
are
equal.
drawn
124
GEOMETRY
PLANE
Proposition
chords
are
circle
with
0
be the distances
OH
To
the
circle, if two
same
chord
is at
the
less distance
the center.
from
Given
Theorem
equal circles, or in
unequal, the greater
In
308.
VIII.
OF
prove
chord
AB
and
chord
AE,
of
AB
CD
CD, and
chord
"
from
center
0,
let
OF
and
respectively.
OH.
"
Argument
1.
From
A
draw
a
2.
From
0
draw
OG"AE.
3.
Draw
FG.
4.
AB
CD.
5.
.'.
6.
F
"
AB
"
and
AE,
mid-pointsof
respectively.
G
are
.-.
^F
8.
.-.
Zl
"Z2.
9.
Z
^0
=
10.
.-.
Z3
11.
.'.
OF"
12.
06?
13.
.'.
AB
and
Z
0"4.
"Z4.
0G.
OH.
=
0F"
equal
the
AG.
"
OH.
CD."
Q.E.D.
The
Note.
made
its
DC.
AE.
7.
309.
equal to
;
student
thus, reason
should
5 above
give
the full statement
should be
;
"
of the
SubstitutingAE
stitution
subfor
IX.
Proposition
125
II
BOOK
Theorem
(Converseof Prop. VIII)
distant
unequally
are
less distance
proof is
AB
left
chord
the
at
greater.
with. OF, the distance
chord
prove
The
0
circle,if two chords
same
the center, the
from
OH, the distance
0, less than
To
is the
circle
Given
in the
equal circles,or
In
310.
as
of chord
chord
"
CD
from
AB
from
center
center
O.
CD.
exercise
an
of chord
for the student.
Begin with A OGF.
Hint.
311.
Cor.
I.
A
is
diameter
greater
than
other
any
chord.
312.
Cor.
circle
of a
the
same
the
Ex.
Ex.
locus
The
n.
equal
to
a
center
of the mid-points of all chords
ing
given chord is the circumference havas
the
given circle,and
the center
perpendicular from
433.
Prove
434.
Through
Prop.
a
IX
having for
to the
dius
ra-
given chord.
by the indirect method.
given point within
a
circle construct
the
mum
mini-
chord.
Ex.
435.
If two
chords
making unequal angles with
Ex.
an
436.
are
drawn
it,the
from
chords
one
are
extremity of
a
diameter,
unequal.
The
perpendicular from the center of a circle to a side of
inscribed equilateraltriangleis less than
the perpendicular from
the
of
center
the circle to
a
side of
an
inscribed
square.
(See " 308.)
126
PLANE
GEOMETRY
Proposition
313. A tangent
drawn
to
to the
to the
to the radius
circle is
a
M
T
line AB, tangent to circle
Given
Theorem
perpendicular
point of tangency.
A
To
X.
B
T, and
O at
OT,
a
point of tangency.
"
AB
prove
OT.
Argument
1. Let
be any
M
then
2.
radius drawn
Draw
point on
is outside
M
Reasons
other
AB
than
T;
1.
" 286.
2.
" 54, 15.
3.
" 54, 12.
4.
" 279, a.
5.
" 309.
6.
Arg. 5.
7.
"165.
the circumference.
the circumference
OM, intersecting
at s.
3.
OS
"
4.
OS
=
5.
.-.
OT"
6.
.-.
OT
om.
OT.
OM.
is the
drawn
7.
/.
from
OT"AB
314.
shortest
Hint.
suppose
to
Q.E.D.
a
radius
A
straight line
extremity is tangent
at its outer
circle.
Prove
that
Cor.
of tangency
the
by
AB
through T, tangent
315.
OT.
(Converse of Prop. X).
I.
perpendicular
to the
be
can
to AB.
0
; i.e. AB"
Cor.
line that
II.
is not
to
indirect
tangent
circle O.
method.
to
circle O
at
figure for Prop. X,
point T; then draw CD
the
Apply " 63.
A perpendicular to
passes
In
through
the center
a
tangent
of the
at the
circle.
point
BOOK
316.
A
m.
Cor.
perpendicular
to
line drawn
the center
from
tangent
a
127
II
the
through
passes
of
circle
a
point of
tangency.
317.
of the
Def.
A
polygon
about
a
circle
is
polygon
the
In
circle.
is circumscribed
if each
the
to
tangent
figure the
same
is said to be inscribed
circle
side
in
the
polygon.
The
437.
Ex.
sides of
a
perpendicularsto
circumscribed
points of tangency
point.
Ex.
438.
to the
center
If two
Ex.
441.
a
mon
com-
the
tangents
of
vertex
any
the
a
are
them
circumscribed
that
angle at
adjacent points
bisectors of the
The
through
a
from
to
angle between
440.
at their
circle bisects
radii drawn
bisector of the
pass
line drawn
of the
439.
Ex.
pass
The
angle between
Ex.
polygon
through
the
drawn
passes
from
angles of
a
also
the
of tangency.
point
a
through
polygon
and
vertex
to
a
circle,the
circle.
of the
the center
circumscribed
quadrilateral
point.
common
Tangents
to
a
circle at
the
extremities
of
diameter
a
are
parallel.
Proposition
To
318.
point
The
construct
tangent
the circumference.
in
a
construction,
proof,and
for the student.
Ex.
a
442.
circle.
Ex.
circle.
Ex.
and
and
kinds
443.
What
444.
445.
of
of
Construct
a
given
Construct
'perpendicularto
a
circle at any
a
discussion
are
left
as
given
an
which shall be circumscribed
quadrilateral
?
are
circumscriptible
quadrilaterals
a
Construct
kinds
to
cise
exer-
(See " 314.)
Construct
What
parallelto
Ex.
Problem
XI.
parallelogram which shall be
?
parallelogramsare inscriptible
a
a
inscribed
about
in
a
line which
shall be
tangent
to
a
given
circle
line which
shall be
tangent
to
a
given
circle
line.
a
given line.
128
GEOMETRY
PLANE
319.
The
Def.
included
from
which
of
length
between
to
two
XII.
tangents
Given
To
and
PT
Ex.
is
Ex.
proof is
The
447.
Ex.
the
a
449.
circle is
any
pointP
given point
to circle 0.
equal
as
an
exercise
for the student.
of two
oppositesides
of the
other
sum
median
of
a
two
of
a
circumscribed
eral
quadrilat-
sides.
circumscribed
trapezoidis one
circumscribed
about
fourth
the
trapezoid.
A
448.
or
sum
the
equal to
rhombus
are
tangents from
PS, two
left
The
446.
perimeter of
Ex.
from
equal.
PT=PS.
prove
The
Theorem
drawn
are
cirele,these tangents
a
the segment
point of tangency and the point
tangent is drawn j as TP in the followingfigure.
the
If
lengthof
the
Proposition
320.
is the
tangent
a
parallelogram
a
circle is either
a
square.
The
hypotenuse
to the
sum
of the
of
a
right trianglecircumscribed
other
two
sides minus
a
diameter
about
a
of the
circle.
450.
from
If
the
a
circle is inscribed
in any
and if three triangles
triangle,
cut
are
given triangleby drawing tangents to the circle,then
the sum
of the perimeters of the three triangleswill equal the perimeter
of the given triangle.
Ex.
130
PLANE
GEOMETRY
Proposition
To circumscribe
323.
A
Given
The
Problem
circle about
a
given triangle.
a
ABC.
circumscribe
To
XIV.
circle about
a
A
ABC.
discussion
construction,
proof,and
left
are
as
exercise
an
for the student.
324.
Three
Cor.
determine
Ex.
about
the
an
Discuss
triangle;
acute
Circumscribe
Ex.
457.
Given
459.
Ex.
drawn,
and
will intersect
Ex.
460.
the
461.
If upon
of the
chord
Ex.
their
462.
463.
the
a
line
circle circumscribed
an
obtuse
an
isosceles
triangle.
trapezoid.
triangleand
the
radius
of
triangle.
circumscribed
and
sides of any
circles circumscribed
at
a
The
circles of
equilateral
an
two
The
triangleequilateraltrianglesare
the three triangles,
these
circles
about
point.
common
of
segments
a
which
secant
are
between
two
centric
con-
equal.
are
perpendicular
The
circle and
of the
of
isosceles
an
construct
the
quadrilateralpass through
Ex.
center
straight
concentric.
circumferences
Ex.
of
base
inscribed
The
the
circle about
a
to
circle,
circumscribed
triangleare
same
righttriangle;
a
456.
458.
positionof
the
Ex.
Ex.
the
in
not
circle.
a
455.
points
bisectors
of
the
sides
of
an
inscribed
point.
a
common
bisector
of
an
another
point equidistantfrom
arc
of
a
circle is determined
the
by
the center
extremities
of
the
arc.
If two
mid-points with
chords
the
of
center
a
circle
of the
equal, the lines
circle are equal.
are
which
connect
BOOK
TWO
325.
line
The
Def.
circles is called their
326.
CIKCLES
determined
line
of
Concentric
Def.
centers
circles
the
by
two
center-line.
or
circles
are
of
centers
have
which
the
center.
same
Proposition
If
327.
on
131
II
their
two
line
XV.
Theorem
circumferences meet at
of centers, they also meet
-point which
a
in
other
one
is not
point,
M
N
circumferences
Given
their line of centers
To
as
that
prove
M
and
meeting
N
at P,
a
point not
on
OQ.
the
circumferences
meet
at
one
other
point,
R.
Argument
1. Draw
and
OP
Rotate
about
AOPQ
OP
=
R
.-.
is
Also
.\
R
R
.\
is
QP
=
=
both
on
circumference
M
328.
and
Cor.
of venters
meet
N
I.
If
an
axis until
2.
" 54, 14.
3.
By cons.
" 279, a.
By cons.
" 279, a.
Args. 4 and
positionQRO.
of circle M.
a
of circle
bisects their
6.
and
M
7.
6.
i.e. circumferences
at R.
two
5.
N.
N.
circumference
N:
4.
M.
radius
circumference
on
is
as
circumference
on
QR
OQ
radius
a
=
1. "54,15.
QP.
it falls in the
OR
Reasons
Q.E.D.
circumferences intersect,their line
common
chord
at
right angles.
132
PLANE
329.
Cor.
If
II.
only,that point
If
Hint.
they
they
also meet
330.
on
meet
Two
at
within
circles
they have
or
their line
332.
Cor.
they have
Hint.
333.
a
III.
If
two
hypothesis.
be tangent
or
each
to
They
one
circle lies
be stated
each
circles
Concentric.
follows
as
other,their
:
point lies
common
tangent to each other,
tangent line at their point of contact.
common
Def.
line
A
an
Fig.
are
a
belt
circles is called
touching two
if both
circles lie
internal
common
on
the
tangent
an
external
side of it ; the
same
if the two
circles lie
of it.
Fig.
1.
connecting two
of external
in
tangent to
centers.
oppositesides
Thus
the
contradicts
This
Apply " 314.
line is called
on
line of eenters,
Tangent internally.
of
tangent
common
their
on
point
one
of the other.
If two circles are
on
only
" 330, Cor. II may
From
is not
at
point in common.
externally according as one
Tangent externally.
331.
which
said to touch
are
and
one
outside
or
point
a
point (" 327).
tangent internally
are
circumferences meet
their line of centers.
two
in another
Def.
if
other
is
GEOMETRY
common
Fig. 2 illustrates internal
wheels
as
tangents, while
common
in
2.
tration
Fig. 1 is an illusa belt arranged as
tangents.
334.
tangents
common
many
intersect
wholly
Ex.
in
?
464.
If two
circles
from
them
Ex.
at
given circle
a
circle at
are
the other ?
bisects
centers
the
line
5 inches
are
and
joining their centers
?
externally,tangents drawn
to each
tangent
other.
passing through
are
to
equal.
Construct
their
mon
com-
a
given point and
gent
tan-
given point.
of all circles
the locus of the centers
Find
469.
they
case
within
contact.
another
at
wholly
internal
common
circle
a
circumferences
in
points of intersection.
of the
tangent
are
is
tangent
are
point of
their
Construct
468.
Ex.
circles
Two
tangent
to
circles
their
case
intersectingcircles
two
length
point in their
any
467.
Ex.
may
If two
466.
the
be
internallyhow
externally?
one
the
to
radii of
If the
case
tangent
intersect,their line of
radii drawn
the
inches, what
Ex.
? in
other
are
in
?
tangent
are
of each
465.
Ex.
they
case
drawn
be
can
circles
two
case
outside
angles between
8
In
Questions.
133
II
BOOK
tangent to
given
a
given point.
a
MEASUREMENT
335.
To
Def.
it contains
the
another
is
measurement
The
how
a
many
The
as
It
336.
there
the
foot
feet there
a
be
can
proof that
measure-number
to
shown
every
is found
proofthere used shows
follow the ordinarylaws
times
Def.
a
measure
in each.
of.the two
called
The
result
the
numerical
quantitywhich
of the
of
of
is measured.
measure.
is measured
in it ; i.e. tiow many
are
that
to
every
ing
find-
by
it
times
tains
con-
Two
straightline segment
Such
quantities.
belongs a
that
of
The method
Appendix, " 595.
with measure-numbers
operations
algebra.
are
is contained
a
there
in the
quantities
that
geometric,magnitude
called its measure-number.
definite number
of
exists
is
times
many
measure.
corresponds a
337.
and
number
employed is called the unit
length or breadth of a room
measure
Thus,
a
measure-number,
or
measure,
how
quantityis to find
kind.
quantityof the same
a
measure
measure
commensurable
an
is called
if
there
integral number
a
common
measure
of
134
PLANE
Thus,
inch
an
whole
a
inches
18f
inch
What
If two
18|
is their least
and
numerical
a
Name
common
some
an
how
measure,
common
greatest
of
measures
common
?
measure
there
if
incommensurable
is contained
later that
integralnumber
an
of
that
the
of
measure
1.4 and
lies between
Again, it lies
between
values
between
1.4142
to
V2, but
as
a
and
be
may
found
decimal
no
diagonaland
a
side of the
a
same
mainder;
unit, without a reby the same
diagonal is equal to V2 times the
V2 can be expressed
the
side.
Now
measured
only approximatelyas
a
simple fraction
or
as
a
decimal.
It
1.5, for (1.4)2 1.96,and (1.5)2 2.25.
1.414 and 1.415,
1.41 and 1.42,*between
=
=
1.4143, and so on.
approximating more
number
By repeated
and
be obtained
can
trials
closely
more
that,taken
twice
factor,will give exactly2.
340.
When
the
some
or
has
numbers
two
in mind
between
difference
the other
of
speak
we
have
another,we
been
the
341.
defined
as
measure
*
The
the
defined
student
many
By
quantity to
this is meant
times
one
tains
con-
algebrathe ratio
the indicated
as
quotient of
Since to each geometric magnitude
aliquotpart
ratio
of
quotient of
is
one
of it.
In
called
its measure-number
:
The
Def.
of
the two, but how
by the second.
correspondsa number
(" 336),therefore
ratio
their relative sizes.
first divided
there
same
of
?
measure
that
be
cannot
square
the
?
quantitiesare
It will be shown
of
and
in each.
times
not
fourth
containinga
is their
What
common
measure
no
they
inches.
Two
Def.
exists
each
containing
too, 12" inches
so,
;
quantitieshave
have
measures
and
339.
of times
each
of times.
Questions.
inches
commensurable,
are
number
number
common
many
foot
a
commensurable,
are
whole
a
338.
12"
yard and
a
GEOMETRY
appliedto
two
geometric
their
magnitudes
may
measure-numbers, when
each.
should multiplyto get the successive approximations.
be
the
Thus, if the length of
the ratio of the
is
length
applied to
never
342.
the
ratio
is
If the two
Def.
the
be
to
feet,
ratio of
ratio of the width
The
ratio is
term
unlike.
commensurable
a
27
the
magnitudes compared
is called
a
the width
.
two
always be expressed as
343.
is said
equal to f The
which
is equal to j.
|-J,
two
magnitudes that are
the
If
Def.
the width
length to
36 to 27 ; i.e. ff, which
to the
is 36 feet and
room
a
135
II
BOOK
are
surable,
commen-
and
ratio
can
simplefraction.
magnitudes compared
ratio is called
incommensurable
an
surable,
incommen-
are
and
ratio
can
be
expressedonly approximatelyas a simple fraction. Closer
incommensurable
ratio may
and closer approximations to an
be obtained
by repeatedlyusing smaller and smaller units as
of the two magnitudes to be compared and by finding
measures
the
quotientof
thus
the numbers
obtained.
are
magnitudes,e.g. two line segments, taken at random
usuallyincommensurable, commensurabilitybeing comparatively
Two
rare.
344.
Historical
is ascribed
discovery
that
complete
18
Ex.
Show
also
the
Ex.
a
a
48
Pythagoras
common
"
magnitudes
kept the
time
the first to prove
in
A
inches
?
which
have
48
more
" 510.
of 48
measure
18
long
was
will be found
Pythagoras
inches
a
incommensurable.
are
square
of
473.
and^point out
two
the
lines
that
the
and, by laying off
inches
inches
and
2 x
"
18
any
inches
common
plus or
greatest
similarityof
and
measure
minus
common
the
the
a
common
ure.
meas-
common
measure,
of the lines.
sum
lines,5
two
of 15 inches
Find
line segments
of the
measure
diagram
measure
any
of these
sum
Given
472.
by
a
a
greatest
it divide
Draw
471.
that it is
Ex.
is the
that
for
?
Find
show
of the work
Will
?
of
of incommensurable
followers
whose
is believed
diagonal
What
470.
discovery
Pythagoras,
It
and
account
inches
inches
to
secret.
side
the
Ex.
a
The
Note.
process
4 inches
long,respectively.
of 5 inches
and
4 inches
844
by
is
8 inches.
divisor of 728 and
to that
used
in
Prop.
division
XVI
136
PLANE
GEOMETRY
Proposition
To determine
345.
or
their
common
whether
and
not;
given lines
two
if they
i-i-j
E
"
and
AB
determine
To
I.
AB
2.
off ED
Measure
on
it is contained
3.
off
Measure
on
off
Measure
GD
5. It is evident
a
remainder
common
AB
that
is obtained
to CD.
AB
remainder
a
as
ED
FB
on
and
measure;
times
many
a
as
a
a
as
possible. Suppose
remainder
as
GD.
and
possible,
will terminate
this process
is
pose
possible. Sup-
as
FB.
times
times
many
which
ED.
remainder
as
possible.Suppose
as
times
many
many
three times,with
it is contained
4.
as
twice,with
FB
commensurable
are
Construction
with
once,
CD
is the ratio of
CD
on
and
AB
is their
(6) what
(c)what
it is contained
i_i
"
if so,
and
off
"
CD.
(a)whether
:
1. Measure
G D
i
i
i
i
lines
find
to
B
F
i
C
Given
surable
commen-
commensurable,
are
A
i
are
their ratio.
and
measure
Problem
XVI.
on.
only when
of the
measure
so
remainder
immediatelypreceding.
terminates,then
6. If this process
commensurable,
the last remainder
and
the
two
given
is their
lines
greatest
are
mon
com-
measure.
7.
are
For
example,if
commensurable,
the ratio of
AB
to CD
GD
GD
can
is
a
measure
is their
of FB, then
greatest common
measure,
Proof
"Reasons
Argument
Suppose
2.
ED
=
EG+
FB
=
GD.
2 GD.
and
be found.
II.
1.
AB
1.
See
2.
"54,11.
I, 7.
CD
and
138
PLANE
Ex.
but
Consider
475.
variable
would
there
Ex.
lines
in this
be
If
476.
drawn
are
348.
A
point
a
A
other
what
the
whose
the
in
equal
perimeter
positionof
is
magnitude
the
base
sides
of
will be
the
of
Def.
If
a
the
constant
triangle
parallelotriangle,a gram
; i.e. the perimeter
point.
if it does
constant
variable
approaches a
between
the variable
difference
be made
angles,
variables
isosceles
an
change
not
if it takes
magnitude is variable
successive values during a discussion.
that the
to become
and
remain
previouslyassigned,however
smaller
in such
constant
a
way
the constant
and
than
small,the
series of
a
may
fixed number
any
is called
constant
the
of the variable.
limit
The
350.
more
and
move
from
is said to approach
variable
more
A
successive
nearlyequal
toward
B, by
that
step it must
go
remains
half the
one
the first step it reaches
to be traveled
over
of
the
segment last covered.
some
value,say,
move
under
remains
value
a
its
pointmay
less than
But
the
F ?
any
an
the
a
D, and
segment equal
segment between
Then
variable
half
an
segment
to differ from
assigned value.
may
the
there
the number
by
AB
as
A
and
little as
to
the
we
For
assign
the point,continuing to
approach B until there
previouslyassigned value.
inch.
At
B.
the segment
Whatever
to differ from
segment less' than
be made
it and
step it reaches
always remain
governing law,
assigned,the
point to
a
^
segment between
to be covered.
be made
half
it becomes
"
; at the next
equal segment
steps taken,there must
ptease, i.e.by less than
suppose
C, whereupon there remains
an
moving point may
Thus,
as
each
at
over
it.
to
its limit
A
the
steps, under
restriction
CB
and
constants
base
constant
discussion.
Def.
different
349.
any
parallelto
Def.
throughout
other
trianglewith
case.
independent of
347.
what
through
will be formed
will be
isosceles
an
Tell
base.
GEOMETRY
inch.
from
constant
Whatever
A
to
the
segment
be
the
moving
AB
by
each
term
limit
as
is
half of the
one
4, 2, 1,^,\,\, etc.,in which
in the series
numbers
Again, the
the number
preceding term, approach
that
of the series
term
no
In
351.
limits
usually such
the
they
that
that
variables
that
attain
cannot
approach
their
limits.
thus,the expression
"
has the limit 1 when
0, and
352.
becomes
x
which
be
may
said
are
to be
353.
Questions.
this
pile forever,how
law
governing
will it become
hour,
a
much
as
fourth
354.
the
add
we
Historical
of
vice
Does
?
for any
yards
gone
the
by
the
front
nearer
ca%you
ten
ran
1000
one
large will
Achilles
known
of
as
fast
start, it could
the
If
he
him
to the
had
covered
; and
tortoise
find the
so
on
but
it
?
One
Tortoise.
of the
during the fifth century
Eleatic
the
as
as
a
School,
these
100
forever
of the
that
tortoise had
tortoise,yet if the
overtaken
:
for, when
:
still be 100
yards,
thus
overtake
fallacyin his argument ?
it would
Achilles
it."
Was
yards
Zeno
if
(say)
Achilles
had
in front of him
still be
would
for
famous
was
be
never
never
value
keep adding
we
pilebecome
the
and
yards, the tortdise would
1000
time
times
lated
re-
versa.
problems involvinginfinite series. Zeno, one
prominent members, proposed this question: He "argued
Achilles
is
are
square
its investigations
of
most
one
variable
one
area
of sand.
bushel
mathematics, founded
Elea, Italy, and
at
?
Note.
schools
a
when
related
depend upon the
If we
add one
quart each hour, how large
quart the first hour, a half quart the second
hour, etc., each hour adding one half as
third
the
quart
for the
large will it become
preceding hour, how
early Greek
b.c,
If
?
the floor is
On
additions
our
value
a
diagonal,and
the
given to
approaches
zero.
depends upon the other so that, if the value of
known, the value of the other can be obtained.
of a
For example, the diagonal and the area
variables,for there is
this
~\~ -L
variables
Two
Def.
approacheslasa;
-
"
X
to
The
are,
kind ;
if
of the
however, variables that do attain their limits.
limitingvalues of algebraicexpressionsare frequentlyof
There
a
For
term
a
as
it is also evident
y-oxro-g-;
0.
become
can
elementary geometry
are
evident
is less than
which
be found
series may
0
in the series is increased.
of terms
it is
assignany value,as y-g-^-g-,
we
139
II
BOOK
get
10
yards
and
nearer
right?
;
in
If
not,
140
GEOMETRY
PLANE
XVII.
Proposition
If
355.
limits
approach as
To
their
limits
are
L
prove
V',which are always equal and
and L\ respectively.
L
L\
=
Ke
Argument
1. Either
L
L',or L=" L\
=
Suppose that one
the other,say
2.
limit
L
L' and
But
4.
.-.
5.
But
.*.
L
356.
V
value
"
L'.
V'.
"
?
and
In the
V
above
student
The
V
the
and
limit
357.
there
V
by
proof are
may
adapt
three
V and
the
3.
" 349.
4.
Args. 2
5.
By hyp.
6.
" 161, 6.
Vf
and
increasingor
argument
above
3.
creasing
de-
to the
XVII
B
H
line CF.
It will be
Note.
are
v'
and
figure,
accompanying
variable
V is represented
where
by the line AB, variable V by
the line CD, limit L by line AE,
to
" 349.
decreasing variables.
are
Apply Prop.
477.
2.
a
q.e.d.
Question.
in which
Ex.
a
assume
is
variables
case
" 161, a.
tween
be-
value
L\
=
1.
V, in approaching
assume
may
j
impossible,since
always equal.
are
6.
a
i.e. V
L
become
V may
this
assume
a sons
L\
"
v' cannot
3.
is greater than
L' ; then
"
L, may
value
proaches
ap-
variables,V and
two
which
if each
equal.
always equal, and
are
limit, then
a
Given
variables
two
Theorem
distinct
variables
that, in
seen
things to
that
be
the
considered
(1)
(2)
Two
The
limits
(3)
The
equality of these limits themselves.
of these
are
two
always equal;
variables ;
applicationof Prop. XVII,
:
Proposition
its
by
angle at the
intercepted arc.
unit Z
(e.
g.
be the unit
To
arc.
measure-number
the
prove
I. If ZAOB
and
(a) Suppose
that
number
of
ZCOE
ZAOB,
of AB, referred
measure-number
to the
equal
circle is measured
a
be
intercepted
by it; let ZCOE
let CE, intercepted
by the unit Z,
AB
and
degree),
a
Theorem
of
center
and
ZAOB
central
Given
any
XVIII.
An
358.
141
II
BOOK
in ZAOB
is contained
Apply Z COE to
Suppose that
AOB
Then
r
-").Now
Reasons
Zaob
1.
" 335.
AOB
2.
" 335.
pose
com-
3.
" 293, 1.
ferred
re-
4.
" 335.
ferred
ZAOB, reunit,equals the
5.
"54,1.
as
ZCOE
a
is contained
in
of Z
is the measure-number
the
measure.
r
to ZCOE
as
equal
central
a
unit.
A
which
ZAOB
the
.\
r
r equal arcs
intercept
circumference,each equal to
is the
measure-number
to CE
5.
.*.
the
as
measure-number
4
CE
a
as
a
unit.
of
AB
on
GE.
unit.
measure-number
to ZCOE
"*
integral
times.
r
referred
4.
an
of times.
Z
2.
ZCOE,
to CE.
Argument
1.
to
commensurable.
are
ZCOE
referred
of
as
of
a
AB,
referred
to
q.e.d.
142
PLANE
I. If Z
(b) Suppose
number
Z
COE
are
that Z
COE
is not contained
and
AOB
The
times.
of
GEOMETRY
commensurable.
is left
proof
in Z AOB
as
integral
an
exercise
an
for the
student.
Some
Hint.
(Why
II.
i
Try
?)
If Z
of
aliquot part
.
\ZCOE,
ZCOE,
Z COE
and
AOB
ZCOE
must
be
a
etc.
are
incommensurable.
Argument
Let
Z
1 be
Z
a
1 to
possible.
Reasons
of Z
measure
Z
of /.AOB.
measure
AOB
as
times
many
will
There
ply
Ap-
COE.
then
be
a
1. "339.
as
mainder,
re-
Z.FOB, less than Zl.
Z
AOF
.-.
the
and
Z
COS
of
measure-number
to Z
CO.E
measure-number
CE
commensurable.
are
as
a
unit.
as
of
a
A AOF,
ferred
re-
unit,equals the
AF}
referred
to
2.
"337.
3.
"
358, I.
143
II
BOOK
Reasons
Argument
take
4. Now
No
smaller
a
Z
is
COE
taken
will
5. Also
as
the
.*.
Z
than the
Z taken
remain
less than
small;
difference
between
AB, less than
the
and
6.
Args.4and5.
7.
"349.
8.
"359.
9.
Arg.
and
signed
previouslyas-
any
however
Z,
the
/AOF
to become
be made
may
" 294.
5.
tercepted
in-
arc
measure.
a
as
between
difference
AOB
the Z
than
measure.
a
will be smaller
FB
appliedas
remainder,
smaller
be
by the
6.
the
/AOB,
to
Z FOB,
it is
"335.
4.
of
measure
a
taken, when
measure
a
small
how
matter
of /.COE.
measure
and
wise
likeand
AF
interceptedby
arc
assigned Z.
AOF
.*. Z
approaches Z AOB
and AF approaches Xb as
the
7.
8.
the
Hence
the
/AOB
limit,and
as
a
of
the
But
of
10.
as
the
variable
as
.40^
of
measure-
a
limit.
of Z
ure-number
meas'
AOF
is
3.
to the measure-number
measure-number
COE
measure-number
359.
the
measure-number
to Z
CE
of Z
AF.
the
.*.
limit.
approaches the
AF
always equal
limit,
a
measure-number
of AB
9.
a
measure-number
approaches
number
as
as
a
as
of
a
ferred
/.AOB, reunit,equals the
of AB,
referred
unit.
as
to
their limit
approaches
will be found
in the
a
then,
limit,
measure-numbers
the measure-number
of the magnitude.
(This theorem
"355.
q.e.d.
If a magnitude is variable and
magnitude varies, the successive
approach
10.
Appendix, " 597.)
of
of
the
the limit
144
GEOMETRY
PLANE
360.
In
Cor.
angles
the
at
have
center
the
the
in
equal circles,or
ratio
same
circle,two
same
their
as
cepted
inter-
arcs.
Hint.
The
the
measure-
the
angles is equal to
Ex.
of the
angles
intercepted arcs.
measure-numbers
of
numbers
ratio of the
the
Construct
478.
their
Therefore
ratio of
the
arcs.
which
secant
a
equal respectivelyto
are
shall cut off two
thirds of
given
a
circumference.
Is the
479.
Ex.
ratio of the
which
arcs
semicircumference
The
361.
the
will be
oc
*
the
it follows
(a) In equal circles,or
measured
by equal arcs
;
( sum
stii/m.
the
a
of
measure
An
Def.
lies
on
Def.
any
*
It
a
is,of
course,
different
student
needs
number
of
number
of its
an
long
means
or
:
circle,equal a7iglesare
multiple of
equal
measure
arcs
angle is said
and
angle is said to be
inaccurate
kind.; but, in
lies
familiar
to become
angle at
to
the
on
the
it has
case,
with
it.
center, referred
interceptedarc
when
anglesis equal
angle is equal
an
the
More
to any
unit
arc
in
its sides
to
to
that
of that
a
segment and
unit
convention
angle, is
arc
its
arc.
accurately,in Prop.
is the
of
segment
a
another
magnitude by
one
become
chords.
in
of the
if its
circle
a
are
inscribed
arc
speak of measuring
this
two
be inscribed
to
the circumference
An
of
of the angle.
sides pass through the extremities
of
is
oc
of the angles.
measures
measure
if its vertex
circle
by.
lengthof.
same
[difference
\ of the
multipleof the
364.
the
conversely,equal
difference
vertex
as
the
the
of
of
measure
the
363.
a
(sum
The
(b)
\p) The
same
(")
macron
directlythat
in
a,ngles.
(c) The
the
using
answer,
for is measured
used
varies
suggests
" 336
equal to
circle
same
Illustrate your
quadrant.
a
oc
in the
?
variation,and
of
From
362.
and
Hence
length'
chords
they subtend
symbol
symbol
of two
ratio
the
so
XVII
same
interceptedby
magnitude
general
that
I, the
measure'
the
measure-
as
the
unit
the
angle.
146
GEOMETRY
PLANE
D
Fig.
All
angles
equal. (SeeFig. 1.)
366.
are
I.
Cor.
Cor.
368.
Cor.
The
locus
of
the
in
inscribed
II.
III.
in
inscribed
Any angle
right angle. (See Fig. 2.)
367.
2.
Fig.
1.
the
segment
same
semicircle
a
of
vertex
right
a
having a given hypotenuse as base is the
having tiie same
hypotenuse as diameter.
369.
than
Cor.
If
480.
degrees
the
circumference
are
whose
Ex.
484.
How
many
each
case
Ex.
arc
greater
24
angle degrees,how
?
how
many
in
the
many
of
rest
angle interceptsan
of
arc
70",how
many
angle ?
there in
degrees are
many
angle of
an
Ex.
Repeat
solutions
of Ex.
485.
intercepted
inscribed
Construct
in it shall be
contains
segment
a
angle inscribed
an
in
a
ment
seg-
is 140" ?
arc
483.
Ex.
the
in
in the
How
482.
Ex.
angle
less
a
(See Fig. 3.)
angle.
acute
an
inscribed
an
there
are
in
angle
tri-
ference
circum-
segment
(See Fig. 3.)
angle.
a
?
If
481.
degrees
an
there
arc
Ex.
is
semicircle
a
obtuse
an
in
inscribed
angle
angle inscribed
Any
V.
Cor.
Ex.
is
semicircle
a
370.
than
Any
IV.
is
484
The
are
any
:
of
segment
(a)
60"
;
to
circle
(6) 45"
483, using
there
a
a
;
case
an
angle inscribed
(c) 30".
given line
each
that
so
of
as
Ex.
chord
483 ?
of the
how
segment.
many
to
?
opposite angles of
an
inscribed
quadrilateralare
plementary.
sup-
BOOK
If the diameter
486.
Ex.
487.
Ex.
By
and
hypotenuse
and
hypotenuse
371.
the
that
all
diameter
of
circle
a
the
of
a
circle construct
a
right triangle,given the
the
righttriangle,
having given the hypotenuse
and
Thales
(640-646 .a), the
founder
of the
base,
as
the circumference,
on
their vertex
Thales
Seven
much
Wise
business
and
sagacity,and
for his practirenowned
cal
and political
ability. He
shrewdness
was
in
Egypt
his
went
to
and
while
there studied
and
astronomy.
story is told that
ditch ;
doing
one
youth,
day
etry
geomThe
while
whereupon
in the
an
old
heavens, when
According
Egypt from measurements
shadow
Thales
of
so
proved that
as
the
not
shadow
thou
seest
not
How
what
computed
thus, "Placing
pyramid, you
the height of
made
by the
pyramid
the
know
thou
canst
is at
the
of their shadows.
is addressed
the
said,"
woman
Plutarch, Thales
to
which
Thales
the stars, he fell into
viewing
a
righttriangle,given
triangleshaving
had
He
Men.
a
ered
discov-
angles right angles.
of
circle construct
a
have
have
one
triangle.
a
Note.
their vertices
was
of the
an
matics,
school of mathe-
is said to
with
base
sides of
hypotenuse.
Historical
earliest Greek
a
the
equal
adjacent angle.
an
altitude upon
the
means
Construct
489.
Ex.
of the
one
will bisect
arm.
By
488.
Ex.
circle is
of
means
an
a
circumference
triangle,the
isosceles
of
147
II
thy
height of
feet ?
is
Pyramids of
gives a dialoguein
Plutarch
the
your
stick at the
same
rays
was
what
"
two
to the
end
of the
and
triangles,
length of
the
stick
of the
This compuof the stick."
pyramid to the shadow
tation
was
regarded by the Egyptians as quite remarkable, since they were
familiar with applicationsof abstract science.
The
Greek
geometry
mind
than 1by
the
of the
Greeks
was
attracted
being more
practicalaffairs
of
in
general ideal
by beauty and
everyday life.
and
by
speculative,the
abstract
relations
148
GEOMETRY
PLANE
XX.
Proposition
372.
line
To
at
construct
perpendicular
a
in
given point
a
(Second method.
Problem
the
given straight
a
line.
another
For
to
method,
"
see
148.)
M
Q"
A
,'P
^'
line
Given
To
and
AB
construct
a
P,
J_ to
a
point in
AB.
at P.
AB
I. Construction
1. With
and
with
at P and
convenient
0, any
OP
radius, construct
as
a
outside
of AB,
circumference
as
center,
cutting AB
'
Q.
2.
Draw
diameter
3.
Draw
PM.
4.
PM
is _1_AB
The
II.
point
QM.
at P.
proofand
discussion
are
left
as
exercise
an
for the
student.
The
the
near
490.
Ex.
without
a
Construct
of each
of the diameters
line
a
when
the
point
P
is at
or
AB.
perpendicular to line AB
its
at
extremity P,
AB.
Through
diameter
is useful
" 372
of the
end
prolonging
491.
Ex.
ends
of
method
one
of the
points of intersection
circle is drawn.
passes
through
the
Prove
other
that
of two
the
line
ences
circumfer-
joiningthe
point of intersection.
To construct
circle
Given
To
and
0
construct
a
a
circle from
ontside
the circle.
to
tangent
pointP
tangent from
a
Problem
XXI.
Proposition
373.
149
II
BOOK
a
point
side.
out-
to circle 0.
P
I. Construction
1. Draw
PO.
2. With
Q, the
radius,construct
of circle
0
3.
Draw
4.
PT
II.
in
a
circumference
pointsT
P T
and
mid-pointof
and
center,and with
as
QO
as
the circumference
intersecting
V.
PV.
tangents from
PFare
proof and
The
and
PO,
discussion
P
are
to circle 0.
left
as
exercise
an
for the
student.
Hint.
Ex.
492.
base of the
is there
and
OT
Draw
OF
and
apply " 367.
the
triangleabout a given circle,
tion
restricWhat
isosceles trianglebeing equal to a given line.
Circumscribe
isosceles
length of the
the
on
an
base
?
arm
one
righttriangleabout a given circle,
is the least length
of the trianglebeing equal to a given line.
What
possiblefor the given line,as compared with the diameter of the circle ?
Ex.
493.
Circumscribe
Ex.
494.
If
B, the tangents
on
circumference
Ex.
upon
495.
^iebase
a
a
circumference
to B
at
the
M
passes
through
points of intersection
the
center
of
a
circle
of the circles intersect
M.
Circumscribe
of the
an
isosceles
about
triangle
being given.
triangle
a
the altitude
circle,
150
PLANE
496.
Ex.
Construct
a
T
To
construct
a
external
common
tangent
given circles
to two
V
,
circles MTB
Given
GEOMETRY
NVS.
and
external
common
to circles MTB
tangent
and
N
VS.
I. Construction
1.
the line of centers
Draw
with
OL
OM
"
Construct
4.
Draw
6.
Draw
7.
TV
497.
and
by
Make
Ex.
and
center
as
circle LPK.
LPK.
"
MTB
373.
at
T.
a
is
modified
are
left
common
NVS.
as
an
exercise
external
for the student.
tangent
to circles MTB
the
if the
equal ?
are
Construct
498.
second
and
same
How
internal
a
tangent
circles.
Hint.
of Ex.
discussion
Construct
circles
to two
O
188.
circles MTB
to
proof and
method.
common
"
tangent
NVS
Ex.
with
TV.
is
Ex.
two
Then,
QN.
QP from point Q to circle
prolong it to meet circumference
QVWOT.
The
method
"
OQ.
radius
tangent
OP
II.
and
OM
$7Vas radius,construct
"
3.
5. Draw
Q
radius
Suppose
2.
Follow
496
OL
except step 2.
OM
=
499.
toward
steps
By
0
+
moving
precedin figure,show
only three ; when only two
in the
tangents ; when
QN.
when
; when
there
only
are
one
four
common
; when
none.
BOOK
Proposition
With
374.
line
with
a
segment
VI
/
\^ ^*"*^"^^\
C
M.
AB
of
segment
a
Z
and
AB
construct,
chord,
*xC
B
Given
Problem
a
4
To
XXII.
given line as chord, to construct
capable of containing a given angle.
circle
of a
151
II
S^XH
^
as
/
circle
a
capable of containingZ
]"
Jf.
I. Construction
1.
2.
AB
" 125.
ZCDE, equal to the given Z M.
With
point on DE, as center and with
H, any convenient
point,as K.
radius,describe an arc cuttingDC at some
Construct
as
3.
Draw
4.
Circumscribe
5.
Segment
II.
The
375.
AB
as
Z,
an
UK.
a
proof and
and
376.
a
discussion
Z
vertex
given base and
which
forms, with
a
arc
of containing
Ex.
angle of
Ex.
105"
501.
base
Ex.
On
500.
;
a
Z
of
a
What
M?
can
is the
the vertices
given angle
of
the
line
construct
a
of the
sum
all
a
segment
a
with
A
A ?
base
ing
triangles hav-
their
at
given base,
given angle.
given
construct
you
vertices
segment
that
is the
capable
shall contain
an
of 135".
Find
2 inches
502.
the
left to the student.
are
moving AB,
=
locus
" 323.
AKDH.
segment capable of containingZ-M.
Without
The
Cor.
circle about
is the
KDH
Questions.
base
as
the
long
Construct
locus of the vertices of all
and
a
having
their
vertex
triangleshaving a
angles equal to 60".
triangle,having given ", hi, and
B.
mon
com-
152
PLANE
GEOMETRY
XXIII.
Proposition
An
377.
within
by
angle formed
circle is measured
a
tJie sides
of
by
To
chords
two
AB
and
that Zl"i
prove
the
arc
of the arc
tween
intercepted besum
B
at
CD, intersecting
Reasons
" 54, 15.
AD.
Z1=Z2
+
Z3.
"215.
Z_2*\Bb.
Z 3
"365.
\2b.
*
E.
(BD + AC).
Argument
Draw
intersect
angle.
tL
Given
half the
one
and
its vertical
which
chords
two
its sides
between
intercepted
Theorem
"365.
" 362, ".
.'.
Z 1
Ex.
arc
oc
i
(BD
One
503.
of 40".
AC).
+
angle
Its vertical
"309.
Q.E.D.
formed
by
intersecting chords
two
angle intercepts an
of 60".
arc
How
interceptsan
largeis the
angle ?
Ex.
If
504.
is
arc
Ex.
505.
the
sum
angle of
an
30", how
large is
the
oppositearc
is 40"
of two
intersect
and
cepted
its inter-
?
right angles within
oppositeinterceptedarcs is equal to
chords
If two
intersectingchords
two
at
a
a
ence,
circumfer-
semicircum-
ference.
Ex.
if AM
at
D,
506.
is
If M
prolonged
prove
that BD
is the center
to meet
ss
DM
the
=
of
a
circle inscribed
circumference
DC.
of
the
in
triangleABC
circumscribed
and
circle
154
PLANE
GEOMETRY
Proposition
An
XXV.
Theorem
angle formed
side
outby two secants intersecting
of a circumference, an angle formed by a secant and
a
tangent, and an angle formed by two tangents are each
measured
by one half the differenceof the interceptedarcs.
379.
Fig.
An
I.
angle formed
two
Given
To
by
secants
that Z
prove
A
cc
Fig.
BC,
(AC
forming
Z 1.
BE).
"
Argument
Draw
3.
.-.
4.
Z 3
5.
.-.
Z3-Z2
6.
.-.
Z 1
Z 1
The
Z
3
"215.
Z 2.
-
ac, and
-i-
An
1
(AC
-
Note.
(1)
:
\
" 365.
BE.
BE).
II and
In
within
the
the
Q.E.D.
by
angle formed
proofs of
"
\{ac-be).
"*
QC
" 54, 3.
Z 2
angle formed
An
380.
be
Z3.
=
=
cc
III.
may
" 54, 15.
+ Z2
Z1
II.
Reasons
CD.
2.
3.
(Fig.1).
secants
and
BA
1
two
2.
a
by
III
secant
a
"362,6.
6.
" 309.
tangent (Fig.2).
tangents (Fig.3).
two
left to the student.
are
preceding
circle ;
and
5.
(2)
theorems
on
the
the
vertex
circumference
of
;
the
(3)
angle
outside
the circle.
Ex.
the
511.
three
Tell how
to
measure
possiblepositions with
angle having its vertex
regard to the circumference.
an
in
each
of
XXVI.
Proposition
lines
Parallel
381.
155
II
BOOK
Theorem
intercept equal
1.
Fig.
IIlines
I. If the two
Given
To
IIchords
AB
AC
BD.
prove
=
and
H
2.
secants
are
or
chords
(Fig.1).
arcs
CZ), intercepting
CB.
1.
" 54, 15.
Z2.
2.
" 189.
3.
" 365.
4.
" 362, a.
5.
"
Z1
3.
Z.\^\ac, andZ2ocij"b.
.-. \AC=\BB.
4.
5.
AC
.'.
If
II.
BD.
Reasons
2.
=
and
AC
Argument
1. Draw
ference.
circum-
a
on
"
n
Fig.
arcs
i"d.
=
one
Q.E.D.
of the IIlines is
a
and
secant
the
54, 7
other
a
a.
tangent
(Fig.2).
III.
If the two
The
proofs
of
IIlines
and
II
tangents (Fig.3).
are
III
left
are
as
exercises
for
the
student.
Ex.
In
512.
Fig. 1, Prop. XXV,
degrees are there
many
equals 40", how
Ex.
the
of
number
Ex.
514.
Ex.
515.
minus
Ex.
the
516.
Fig. 2, if arc DC equals 60"
degrees in angle 1.
In
513.
angle 1 equals 42" and
in angle 3 ? in arc AC?
if
In
angle
interceptedarc.
If two
arc
formed
tangents
do
by
to
a
two
tangents
circle meet
they intercept?
BE
angle 3 equals 100",find
Fig. 3, if angle 1 equals 65", find angle 2
An
degrees of
nxaftjr
and
arc
at
and
angle 3.
is measured
an
angle
of
by
180"
40", how
156
PLANE
Ex.
517.
Is the
converse
Ex.
518.
If two
sides
other
sides
two
of
of
are
equal.
converse
519.
Is the
Ex.
520.
If, through
the
522.
Prove
that
Ex.
523.
If two
pairs of
sides
two
of
arcs
diagonals AC
the
in
there
of
parallelto
the angle.
inscribed
an
side of the
one
AD
arcs
trapezoid inscribed
a
sides of
in
circle is isosceles.
a
inscribed
an
hexagon
parallel,
are
equal.
are
sides AB
If the
524.
subtend
is drawn
answer.
your
bisector
the
parallel,
through the points of intersection of two circumferences
drawn
terminating in the circumferences,these parallelswill
Ex.
Ex.
the
answer.
your
If
are
parallels
be equal.
the other
Prove
?
side of
other
Prove
quadrilateralare
where
chord
?
true
518 true
point
circumference,a
chord
will equal the
XXVI
inscribed
an
the
521.
Ex.
Prop.
of Ex.
Ex.
angle cuts
angle, this
GEOMETRY
and BC
of
inscribed
quadrilateralABGD
if angle AED,
and
formed
60" and 130",respectively,
by
and BD
intersectingat E, is 75",how many
degrees are
and DC
? how
rilateral
degrees in each angle of the quadmany
an
?
EXERCISES
MISCELLANEOUS
Ex.
Prove
Ex.
Equal
525.
that
CE
line of centers
the circles.
Ex.
Through
draw
a
to
from
;
P
tangent is drawn
second
a
circles intersects their
unequal
two
at E.
chords.
the
by
intersect
is C
center
to
one
of
that it is also tangent to the other.
Prove
527.
point P
a
circle whose
a
tangent
common
at
of
angle formed
bisects the
A
526.
chords
of the
one
chord
of
points
intersection
shall
that
one
of
of two
bisected
be
ferences
circum-
the
by
other
circumference.
Ex.
of
a
The
528.
circle ; AC
angle ABC
inscribed
is any
is prolonged to P,
CP
making
angle
equal to
in
CB.
a
given segment
the locus
Find
of P.
Ex.
Find
Given
529.
the
locus of the foot
the latter revolves
Ex.
the
at its ends
531.
around
If two
530.
point of
Ex.
two
are
contact
points P and Q, and a straightline through Q.
of the perpendicular from P to the given line,as
Q.
circles touch
and
each
terminated
and
other
by
the
a
line is drawn
circumferences,the
through
tangents
parallel.
If two
circles touch
each
other
and
two
lines
are
the
the circumferences,
through the point of contact terminated by
joiningthe ends of these lines are parallel.
drawn
chords
other
circles touch
touches
from
distances
cuts
of
the
circle,
hypotenuse bisects
the
a
the
other
each
Show
externally.
both
of the variable
center
externallyand
that
the
circle to the
circle
a
difference
of
of the
centers
circles is constant.
fixed
If two
534.
Ex.
circles
tangent bisects their
Ex.
tangent
each
external
circles
drawn,
points of
circles to the
tangent.
lines drawn
from
common
point of contact
the
of the external
contact
if their
tangent externallyand
are
internal
common
tangent
of
the
perpendicular
are
other.
The
536.
Ex.
is
their
tangent externally,
are
common
If two
535.
external
to
diameter
righttriangleis the
a
the circumference
fixed
Two
radius
of variable
the
of
arm
arm.
533.
Ex.
one
point where
at the
tangent
the
If
532.
Ex.
157
II
BOOK
line of centers
two
at
a
external
common
the line of centers
meet
at a
the
radii
circles meet
to two
two
their
internal
common
gents
tan-
point.
common
circles whose
Two
Also
point.
common
tangents
17 and
10
inches,respectively,
are
long is the line joining their centers ? how
tangent externally. How
if
circles
the same
are
long
tangent internally?
Ex.
537.
Ex.
538.
If
rightangle at
a
also trisected ?
interceptedarc
the
are
of
center
Is the
chord
circle is
a
which
is
trisected,
subtends
the
the
sected
tri-
arc
?
Ex.
a
that
way
two
Draw
539.
a
chords
the
given lines.
line
intersectingtwo
interceptedby
What
restriction
the
given
circumferences
circumferences
two
is there
the
on
lengths
in such
shall
equal
of the
given
lines ?
Ex.
the
median
Ex.
are
Construct
540.
to the
equal, if
Under
base.
In the
541.
triangle,given
a
sides of
two
one
conditions
what
or
circle,
same
are
its base, the
angle, and
vertex
will there
be
no
solution ?
in
inscribed triangles
two
equal circles,
equal respectivelyto two sides of the
other.
Ex.
two
lines
join their
Ex.
If
542.
are
drawn
extremities
543.
The
rectangle,which
Ex.
hypotenuse
of
a
the
tangents drawn
is not
The
544.
points of intersection of two circumferences
terminating in the circumferences,the chords which
are
parallel.
through
line
a
square,
through
form
a
vertices
of
an
inscribed
rhombus.
joiningthe center of
to the vertex
righttriangle,
righf^angle.
the
the
square
of the
described
upon
rightangle, bisects
the
the
158
PLANE
Ex.
If two
545.
tangents
Ex.
546.
distance
to two
the
external
tangents
circles,the
segments
common
drawn
are
between
GEOMETRY
points of
Through
contact
547.
In
given circle inscribe
a
shall be tangent to another
Find
548.
in
given point
Ex.
having
the
with
locus of the
given
circumference.
the
Z EOF
ABO
of
the
angle of
Ex.
In
120"
diameters
Ex.
The
equiangular
of
and
are
If
centers,
tangent
are
that
circles described
the
P
triangles
arc
which
angle equal
to
the
chord
any
an
the
radius
two
on
the
0 and
sides of
a
triangleas
cumferences
cirtheir
circles
Fig.
Fig.
1.
2.
other.
Q Fig. 1.
of
Proof
II. M
not
between
O and
.-.
QP"OP.
QP"OM.
OQ+
QM.
QP"
.-.
P
Q, Fig. 2.
0Q+
.*.
on
two
upon
the
OM.
is not
a
third side.
.'.PQ"MQ.
.'.
O
is the
equilateraltriangle,having given
OP+PQ"OQ.
OP=
and
vertex
from
mutually
equal.
each
to
between
A
drawn
In
Outline
M
chord
triangles
meet
line
a
the altitudes of
the
rightangles subtends
an
upon
about
554.
Ex.
at
two
All
circle
of
of
circle.
circumscribed
same
of the
one
longed
pro-
center.
intersect
553.
given length which
the
Construct
552.
of
point
at
a
circle,prove
a
at the
inscribed
Ex.
middle
intersections
altitudes.
radius
a
551.
of the
given
a
quadrilateral
supplement of Z FOE.
supplement of Z BOA.
is the
bisects
be
is
550.
which
at
circle.
given angle
a
segment capable of containing an
given angle at the vertex.
of the
/.EOF
FOEC,
and
base
Let
intersection
locus of the
base
supplement
Ex.
cepted
tangents inter-
parallellines
two
chord
a
given
the
The
given
a
Hint.
.-.
a
549.
forms
these
apart.
Ex.
Ex.
of
internal
common
equal.
are
given points draw
two
two
or
circumference
".
is not
on
circumference
S.
Ex.
is
intersect,neither point of intersection
circumferences
If two
555.
159
II
BOOK
their line of centers.
on
of the
In
556.
Ex.
inscribed
circle
circle tangent to
Ex.
right triangleABG, right-angledat G, the radius
c) and the radius of the escribed
equals \{a + b
any
c
equals |(a
In
557.
"
the
accom-
panying figurea, b, c,
sides of triangleAB G.
1. a
b +
+
c).
^^^~
the
are
~"\
jT
^v"(
Prove:
^N.
s*y
1$T
\
Sr^f
TP;
=
B P
T
.
\
2.
AP=l(a
+ b +
3.
TB
+
Ex.
centers,
so
a
the
shall touch
semicircumference
a
the
within
other
two
the vertices
other
given
a
and
a
;
of
a
circumference.
given triangleas
two.
circle three
also the
so
equal circles,
given circle.
ha, hc, a
561.
Ex.
562.
A, B, and B,
Ex.
563.
a,
Ex.
564.
C and
Ex.
565.
C
Ex.
566.
r, the
radius
the
of
the
circle.
circumscribed
B, B.
the
and
segments
the
segments
of the
radius
of
of
made
c
c
by tc.
made
by hc.
circle,and
inscribed
the
segments
by tc"
Construct
a
righttriangleABC,
Ex.
567.
c, hc.
Ex.
568.
c
Ex.
569.
The
segments
Ex.
570.
The
segments of
Ex.
571.
c
and
a
Ex.
572.
c
and
the
Ex.
573.
c
and
the distance
Construct
that
triangle,
having given :
Ex.
made
;
circles about
Construct
shall touch
Construct
c
each
that
560.
Ex.
quadrant
a
Describe
559.
""\
c-b).
Trisect
558.
Ex.
each
l(a
=
/
^*^-_^/
c);
a
and
square,
Ex.
574.
The
Ex.
575.
A
Ex.
576.
The
segment
one
of
c
c
right-angledat C, having given :
of
made
made
line I, in which
c
made
by hc.
by hc.
by tc.
the vertex
perpendicular from
from
G to
vertex
a
having given:
diagonal.
of
a
diagonaland
a
side.
must
G to
point P.
perimeter.
sum
of G
a
lie.
line I.
of
160
PLANE
Construct
rectangle,having given :
a
non-parallelsides.
Ex.
577.
Two
Ex.
578.
A
Ex.
579.
The
Ex.
580.
A
diagonal and
Ex.
581.
A
side and
Ex.
582.
The
Construct
GEOMETRY
side and
diagonal.
a
perimeter and
diagonal.
a
angle formed
an
angle formed
an
perimeter and
side and
Ex.
584.
The
perimeter and
Ex.
585.
One
angle and
Ex.
586.
The
two
Ex.
587.
A
side and
the
sum
Ex.
588.
A
side and
the
difference
diagonals.
the
by
a
diagonal.
a
diagonal.
a
diagonals.
of the
diagonals.
of the
diagonals.
parallelogram,having given :
a
non-parallelsides and
Ex.
589.
Two
Ex.
590.
Two
Ex.
591.
One
side and
Ex.
592.
The
diagonals and
an
non-parallelsides and
isosceles
the
angle.
an
diagonal.
a
diagonals.
two
angle formed
an
by them.
trapezoid,having given :
and
diagonal.
Ex.
593.
The
bases
Ex.
594.
The
longer base, the altitude,and
Ex.
595.
The
shorter
Ex.
596.
Two
a
diagonals.
diagonal.
A
Construct
diagonals.
the
rhombus, having given :
a
583.
Construct
the
by
angle formed
an
Ex.
Construct
by
a
base, the altitude,and
their included
sides and
angle.
trapezoid,having given :
and
the
angles adjacent to
base.
597.
The
bases
Ex.
598.
The
bases,
Ex.
599.
One
base, the
Ex.
600.
The
bases,a diagonal,and the angle between
a
equal sides.
of the
one
Ex.
Construct
equal sides.
of the
one
the
circle which
altitude,and
two
shall
their included
angle.
the
diagonals.
:
601.
Touch
a
given circle
Ex.
602.
Touch
a
given line
Ex.
603.
Touch
three
Ex.
604.
Touch
Ex.
605.
Have
and
at P
pass
through
a
given point Q
I at P.
given lines
given line I
its center
angle.
diagonals,and
Ex.
a
an
one
two
at P
in line
of which
and
Z,cut
are
also touch
I at
P, and
parallel.
another
touch
a
line
m,
circle K
162
the
Proposition
I.
numbers
is
extremes
Given
To
GEOMETRY
If four
388.
of
PLANE
a
b
:
ad
prove
proportion, the product
the product of the means.
to
d.
c :
=
in
are
equal
Theorem
be.
=
Argument
1
Reasons
"
"
-
b~
2.
bd
3.
.'.ad
is
389.
Note.
merely
the
equations
in
may
the
product
390.
of
of
one
the
to the
Ex.
(3) for
Ex.
to
144 and
Ex.
=
bd.
bc.
student
q.e.d.
should
observe
By hyp.*
2.
Byiden.
3.
"54, la.
that the
used
process
of the
simplifiedby placing the product
as
so,
means
here'
fractional
also, proportions
equal to
extremes.
The
I.
proportional between
mean
to the
their
of
root
square
two
bers
num-
product.
proportions have any three terms
to the three corresponding terms
equal respectively
other, then the remaining term
of the firstis equal
Cor.
If
II.
remaining
606.
607.
608.
Find
two
of
term
the
Given
(4) for
m,
the
equation
m
second.
: r
=
d
: c
; solve
(1) for d, (2) for
r,
c.
proportionalto 4, 6, and
the fourth
10 j to
4, 10, and 6;
4.
Find
the
Find
the
proportional between
mean
9
144
and
; between
9.
609.
392.
in Ex.
be
of the
equal
10,6, and
Ex.
=
1.
d
algebraic "clearing of fractions,"and that
algebra are usually simplifiedby this process,
Cor.
is
391.
The
'
607
third
proportionalto f
What
Questions.
rearrangement
without
the
affecting
required term
Ex.
610.
Find
the
Ex.
611.
Find
the
*
third
and
of
proportional to
See " 382 for the three wdys
of
; to
62 and
"
a2
f and f.
numbers
? in Ex.
proportionalto a2
fourth
f
"
can
? in Ex.
608
a
b2,a"
writing a proportion.
be
"
made
609
?
b.
b, and
a
+
6.
proportion
If in any
612.
Ex.
consequents
equal
are
Ex.
613.
If
Ex.
614.
If I
Ex.
615.
If
x
:
Ex.
616.
If
x
:
a
b
the
that
:
d,
=
b
:
f", prove
y
=
b
:
c, prove
y
=
6
:
c, is (fcc: y
k
=
antecedents
equal, then
are
the
conversely.
c
:
:
and
163
III
BOOK
prove
ma
that
that
Ir
kr
=
cd
a
:
II.
Proposition
:
kb
dx:y
6
=
:
mc
=
:
be
=
kd.
: mc.
bd:c.
proportion?
true
Theorem
(Converseof Prop. I)
"
is equal to the prodIf the product of two numbers
uct
either
the
be made
pair may
of two other numbers,
the other pair the extremes
and
of a proportion.
393
means
ad
Given
To
be.
=
a:b
prove
c:d.
=
Reasons
1.
c
a
=
-
b
394.
Note.
The
be
chosen
must
of the
ber
and
as
so
Ex.
617.
Given
pt
Ex.
618.
From
the
r:l
can
619.
" 54, 8
if hi
M
K
=
:m
Form
m:s,
=
=
l :s,
a
Prove
s:l
s
:m
m:r,
=
=
l:r,
proportion from
the proportionsobtained
be
to
I
c
the
that
prove
in
Arg.
2
first memh
-
k
=
-,
we
I
=
c
:
t ;
also,c
Im, derive
=
and
b
=-.
: r
p
wish
we
f
equation rs
,77
?
i.e.
quotient in
desired
kf, and
divisor
the
that
observe
fl
cr.
=
=
and
means
a.
for the student.
give the
to
as
the
be made
should
fl
Ex.
q.e.d.
exercise
an
pupil
; then
by fl
r
c:d.
=
d may
equation : thus,
divide
must
a
is left
the extremes
3.
i
a:o
d
proof that
The
above
t
i.e.
;
-
2.
By hyp.
By iden.
m:r
=
s:l 7
=
r
:
m:s
=
r
x
4
=
verified ?
m,
3 x
a
r.
following eight proportions:
s:m,
s
7
t:
=
=
:r
I:
the
:p
; from
:l.
ft
=
gb.
How
164
PLANE
GEOMETRY
Ex.
620.
Form
a
proportion from
(a
Ex.
621.
Form
a
proportion from
m2
Ex.
622.
Form
a
proportion from
c2 + 2 cd + d2
Ex.
623.
Form
making
a;
(1)
an
624.'
Ex.
proportion from
extreme
; (2) a mean.
If 7
+ 3 y
:
12
2a
=
+ y
Proposition
c)(a
"
2
(a
a
a
+
:
mn
+
n2
=
de.
=
ab.
=
".
+
a
b)(a" 6) =4x,
+
3, find
III.
b)
"
the ratio
a
:
y.
Theorem
in proportion, they are
in
are
If four numbers
is to the
term
that is, the second
proportion by inversion;
firstas tlxe fourth is to the third.
395.
a:b=c:d.
Given
To
b
prove
: a
d
=
:
c.
If four numbers
proportion by alternation;
396.
the
third
a:b
Given
To
prove
c:
=
a:c
625.
If
Ex.
626.
Transform
place in the
x
:
proportion, they
is, the first term
are
in
is to
fourth.
d.
b:d.
"
Ex.
that
is to the
the second
as
in
are
-1/
=
proportion.
y
:
a
is the
|, what
:x
=
4
:
3
so
value
that
x
of the ratio
shall occupy
x
:
y ?
in turn
every
book
in
Many transformations
followingmethod:
397.
the
This
the
method
to
398.
is illustrated in the
If four numbers
proportion by
two
of
the
last two
Given
a:b
To
the
the
reverse
steps of (1).
in
proportion, they are
is, the sum
of the first
that
third
is to the
the
as
sum"
{orfourth) term.
c:d.
=
{a)a
(6)a
:
prove
hypothesis.
first {or second) term
terms
simplest
Theorem
in
are
in its
analysisof Prop. V.
V.
composition;
is to
terms
the
hypothesis and
Proposition
easilybrought about by
equation
an
this derive
from
with
(2) Begin
be
may
conclusion
the
(1) Reduce
form ("388),then
165
-+-b
: a
-f b
:
,
b
=
c
4- d
:
c
=
c
-f d
:
d.
;
I. Analysis
(1) The
conclusions
required above, when
simplestforms, are as follows
in their
(a)
1.
ac
+ad
2. Whence
3.
(2) Now
.*.
a:
bc.
+
ad
b
=
be.
c:d.
"
begin with
the
a:b
2.
.-.
3.
6.
be +
2.
Whence
3.
.-.
bd
a:b
ad +
=
be
=
ad.
=
c:d.
=
reverse
bd.
.
the
steps.
Proof
Reasons
c:d.
1.
By hyp.
ad
=
be.
2.
" 388.
ac
=
ac.
3.
4.
By iden.
"54,2.
5.
"393.
.-.ac-\-ad=ac-\-bc;
i.e.
tions
equa-
:
Argument
1.
4.
1.
hypothesisand
II.
{a)
to
(b)
ac
=
reduced
a{c+ d)= c{a-\b).
.-.a
+b:a
=
c
+ d:
(fyThe proof of (6)is left
q.e.d
c.
as
an
exercise
for the student.
166
PLANE
If*
627.
GEOMETRY
^"1.
?,find*-+Z;
=
5
y
x
y
Proposition
VI.
iy four numbers
proportion by division;
399.
in
firsttwo
of the
last two
a:b
Given
To
is to tlw
terms
(a) a
that
proportion, they
is, the difference of
first(or second) term
is to the
terms
"
(b) a
I. The
in
are
third
as
the
are
the
ence
differ-
(or fourth)term.
c:d.
=
:
prove
Theorem
"
b:a
=
c
bib
=
c
analysisis
left
dic\
"
did.
"
exercise
an
as
for the student.
Reasons
Q.E.D.
(b) The proof of (b)is left
If
628.
Ex.
=
-
-,
find ^^
3
y
;
By hyp.
2.
"388.
3.
4.
By iden.
"54,3.
5.
"393.
for the
y
VII.
Theorem
in
are
If four numbers
proportion, they are
that is, the
and
division;
proportion by composition
of the firsttwo terms is to their differenceas the sum
400.
in
sum
student.
X-^UL.
x
Proposition
exercise
an
as
1.
of the last
Given
To
prove
a
two
ib
a
terms
=
ci
is to their
difference.
d.
-\-bia"
b
=
c-\-d:c
"
d.
167
Reasons
i.e. a +
b:a
b
"
c
=
die
+
Proposition
In
401.
of
a
series
as
a:b
Given
To
=
a
prove
of equal
c:d
-f-c +
is to its
e:f
=
e
of
-f g
2.
"398.
3.
"399.
4.
"54, 8
a.
Theorem
the
the
of any number
quents
corresponding consesum
consequent.
g :h.
=
:
By hyp.
q.e.d.
ratios
sum
antecedent
any
d.
VIII.
is to the
antecedents
"
1.
b +
d +
/ -J-h
a:
=
b.
I. Analysis
Simplifyingthe
conclusion
ab -|-be +
The
be +
their
from
the
=
have
ab -f-ad -f
hypothesis.
:
af + ah.
required for the first member
equivalentsfor the second member,
terms
and
bg
above, we
of
may
this
equation,
be obtained
168
PLANE
Ex.
With
629.
hypothesis of Prop. VIII,
the
+
a
ffl
If
630.
Ex.
=
g
-
Ex.
632.
If .a
Ex.
633.
Given
that
result from
b
:
c
=
the
:
5.
find
d,
show
q
Use
Hint.
that
a
:
"
6
a:"
"
11
=
:
+
a
=
Write
6.
a
b
:
c :
=
aei
prove
division ;
IX.
the
proportions
"
of
terms
;
any
k:l.
cgk : did.
=
Reasons
Argument
By hyp.
b~d'
e
c.
Theorem
i:j
d, e:f=g:h,
bfj
:
c:c? +
"
(2) by alternation
(5) by composition and division.
products of the corresponding
of proportions form a proportion.
Given
cl
;
The
number
VI.
Prop.
inversion
(1) by
terms
Proposition
402.
g.
y
(3) by composition ; (4) by
To
=
r-t+g"q
proportion
taking the
*-J"
proveffl-"+
4
2y
prove
d+f=g:h.
+
q
1
*"ll
If
631.
Ex.
.g
=
t
e:b
+
c
ft
"
=
r
GEOMETRY
_g
.
i _k
aei
2.
cgk
i.e. aei
403.
two
Cor.
are
Hint.
=
in
Given
:b
prove
634.
If
Ex.
635.
If
r
a
7
c
in
multiple
proportion, equiequimultiples of the last
are
proportion.
a
Ex.
"54,
Q.E.D.
If four numbers
of the firsttwo and
also
To
cgk : did.
bfj
:
2.
m
__
=
x:y.
:bm
am
: s
:
b
m
=
=
i
3
:
nx
=
t, is fr
4 and
x
:
:yiy.
qs
:
y
=
"
qm
?
-.ft
Prove
your
8 ;9, find the value
of
answer.
ax
:
by.
170
PLANE
In the continued
642.
Ex.
GEOMETRY
proportion a:b
b:c
=
c\
=
d
d:e,
=
prove
that:
'
62'
c
If
Ex.643.
x*:yB
".
find
8:27,
=
e~6**
b*'
d
2/
Ex.
406.
Vm
If
644.
The
Def.
it is divided.
point is
:
Vw
=
:
The
line
it is divided
is
divided
of
and
A C
are
the parts into which
are
internally
the
line.
at
The
B
externally at
It should
The
1
if this
D
segments
point is
and
AD
are
that in either
be noted
if this
segments
1
,
of the line.
C
CB.
AC
is divided
tion
prolonga-
DB.
the
case
the
on
point of division
is
of each segment.
end
407.
straightlines
Two
Def.
the ratio of
line to either
one
line to its
ratio of the other
"" 586 and
591) will
(a)
The
(b)
TJie limit
of the
divided
are
of its segments
is
equal
be assumed
Thus,
if
a;
is
(1) j
is
a
variable.
a
variable
the
by
and
(Appendix.
:
of the quotientof a
divided
to
corresponding segment.
quotientof a variable by
variable
if
proportionally
Prop. XI, II, the following theorems
In
408.
limit
line
a
extremities
the
I
one
of
AB
D
AB
16, find ^.
segments
between
into which
1
a
is
constant
variable
by
variable.
a
constant
a
is the
the constant.
7c a constant
:
K
(2) If
Ex.
the limit of
645.
is divided
In
by D;
the
x
is y, then
the limit of
figureof " 409, name
the segments
into
which
the
AD
-
is
segments
is divided
bv
".
into which
B.
AB
straight line parallel to one side of
the other two sides proportionally.
A
409.
divides
Fig.
A
Given
I.
Theorem
XI.
Proposition
To
171
III
BOOK
with
ABC
AB
AG
AB
AE
line
triangle
1.
IIBC,
BE
prove
If
and
AB
AB
commensurable
are
(Fig.1).
Argument
1.
a
Let
be
AF
and
Reasons
common
a
AB, and
in
AB
AB
is
tained
con-
"335.
2.
" 341.
3.
" 179.
other.
4.
" 180.
equal parts and
5.
" 244.
6.
" 341.
7.
"54,1.
that
suppose
r
of
1.
measure
times
AF
and
in
AB
s
times.
2.
**
Then
=1.
AB
3.
the several
Through
on
4.
These
5.
.-.
O.
s
AB,
as
lines
F, G, etc.,draw
are
AC
is divided
AE
into
AC
_r
.*.
s
pointsof
||BE
into
and
r
division
lines
to each
||BC.
equal parts.
~~
AE
s
AB
_AC
AB~
AE'
Q.E.D.
172
PLANE
II.
If
and
AB
AD
GEOMETRY
incommensurable
are
(Fig.2).
KC
Reasons
Argument
1. Let
be
m
to AB
measure
a
as
as
and
Draw
AD
be
_AK
AD~
" 339.
2.
" 337.
3.
"179.
4.
" 409, I.
5.
"335.
6.
Arg. 5.
7.
" 349.
8.
" 408, b.
9.
Arg.
m.
commensurable.
are
1.
a
||"C.
HK
AH
than
m
times
many
will then
possible. There
remainder, HB, less
as
.4#
Apply
of AD.
measure
a
AE'
,
take
Now
No
matter
AD
is
a
how
be
small
AD.
of
measure
a
it is
applied as
remainder, HB,
to AB, the
smaller
of
measure
taken, when
measure
will
smaller
a
the
than
measure
taken.
6.
.-.
the
may
be made
less
than
.*.
AH
"
AH
and
and
to become
AB
remain
previouslyassigned
any
line,however
.-.
between
difference
small.
approachesAB
approaches
AD
as
"
limit.
a
as
a
limit.
AD
Likewise
the
difference
and
may
be made
A C
remain
less than
any
line,however
between
to become
AK
and
signed
previouslyas-
small.
6.
173
III
BOOK
Argument
10.
\
11.
.*.
approaches AC
AK
approaches
"
as
a
always
equal
J
*
to
AH
is
"
AD
AD~
11.
"408,6.
12.
Arg.
13.
"355.
.
A
Cor.
Q.E.D.
straight line parallel
other
the
to
side
one
of
angle
tri-
a
segments which
figuresfor Prop. XI, AB
sides
two
iu the
-proportional. Thus,
4.
AE
AE'
divides
into
are
:
DB
AE:EC.
=
(1)
Ex.
and
r
c
that
a
If
two
648.
Ex.
inches
the
on
Ex.
are
lines
are
Ex.
12-inch
divided
If
trianglea
have
651.
a
Ex.
to
line
652.
653.
the
A
line
Apply
cuts
b"sV,;(6) through
the
AC
that
=
DB
:
EC.
w,
a
:
triangleare
a
line is drawn
what
b
cut
are
by
lines
if the two
the
12
lel
paralfrom
the
segments will
segments which
any
are
are
it cut
number
any
18-inch
two
if the
two
of the
side of the
side ?
the
parallels,
proportional: (a)
oblique.
point of intersection
parallelto
of
the
medians
of
triangle,this
a
line
sides in the ratio of 2 to 1.
can
be
divided
from
parallelto
also the two
base
:
right, w,
Prove
off 3 inches
given ratio (measured
sides and
Ex.
A
a
side,into
through
two
of
if
lines
AB
d.
cuts
line is drawn
divides the other
:
sides
into
the
at
prove
(2)
other.
b
=
parallel
; (")
650.
Ex.
: c
If two
649.
lines
each
side and
third
to the
vertex
diagram
inches, and
18
and
Prop. XI,
AD:AE;
=
parallelto
also
:d;
AB:AC
In the
647.
are
of
Using Fig. 2
646.
Ex.
by using division.
Prove
Hint.
two
AK
=
410.
" 349.
"
AC
AB
'
=
limit.
10.
AE
But
13
limit.
a
as
"
AE
12.
Reasons
the
the
at
one
but
point into segments
one
which
end).
the bases
of
trapezoiddivides
diagonalsproportionally.
a
the other
proof of Prop. XI to the case in which the parallel
sides prolonged : (a) through the ends of the
vertex.
174
PLANE
GEOMETRY
Proposition
To construct
411.
the
XIT.
Problem
fourth proportional
to three
given
lines.
T
R
lines a, b, and
Given
To
to
proportional
I.
1. From
S
a,
b, and
c.
Construction
R, draw
point,as
any
\L
b
c.
the fourth
construct
\M
"
indefinite lines
two
and
RS
RT.
lay off
lay off
2.
On
RS
3.
On
RT
4.
Draw
5.
Through
6.
FG
II.
RM
RF
=
and
a
=
ML
=
b.
c.
MF.
construct
L
is the fourth
The
proof
and
IIMF.
LG
proportionalto
discussion
are
" 188.
a,
b, and
left
as
c.
exercise
an
for the
student.
412.
other
Could
Question.
the
segments
a,
b, and
c
be laid off in
any
order ?
413.
I.
Cor.
the third
To construct
proportional to
two
given lines.
414.
Cor.
n.
to two
To divide
or
Ex.
654.
Divide
a
Ex.
655.
Divide
a
more
given line
given lines.
given line
given line
a
into segments
into
into
in the
segments
portional
pro-
ratio of 3 to 5.
segments proportionalto 2, 3, and 4.
Ex.
656.
Construct
two
Ex.
657.
Construct
two
Ex.
658.
If a,
b, and
(a)
c
175
III
BOOK
lines,given their
lines,given
three
are
x:a
their ratio.
difference
their
and
given lines,construct
b:c,
=
and
sum
(")
x
=
x
their ratio.
that:
so
"
"
b
Proposition
XIII.
Theorem
(Converseof Prop. XI)
If
415.
straight line divides
a
proportionally,it
is
parallel to
the third
F
A
ABC, and
A
Given
BE
BE
prove
BE
2.
Suppose
and
BC
AE
IIBC.
other
Reasons
either
are
that
some
triangle
"
Argument
1.
a
C
AB
To
of
side.
E
that
drawn
so
sides
two
IIor
not
BE
is not
line
through B,
II.
but
IIBC,
as
that
BF, is
1.
" 161,
2.
"179.
3.
"409.
a.
II
BC.
3.
Then
AB
AC
_
~
AF
AB
AB
4.
AC
But
5.
.-.
6.
This
7.
.'.
BE
416.
AB
AE'
By hyp.
AF=
AE.
"391.
_
~
is
impossible.
" 54, 12.
IIBC.
Cor.
into
to.the third
" 161, b.
Q.E.D.
If
a
straight line
segments
side.
which
Thus,
if
are
AB-.BB
divides
two
sides
proportional, it
=
AE:
EC,
BE
is
of
a
angle
tri-
parallel
is IIBC.
176
PLANE
659.
Ex.
In the
=
and
If
660.
Ex.
EC
Ex.
=
AE
#"
15, AE
=
BC
parallelto
AB
||BC,
=
?
25,
=
If BE
Def.
If the
to
equal respectively
are
angles
said
in the
be
to
two
angles of the
418.
5,
=
Def.
If
angles of
those
said to have
25, and
EC
of
mutually
another, taken
equiangular.
in
sides of
their sides
The
called
ratio of two
homologous
such
called homologous
order,are
in order
polygon, taken
equal ratios with
one
the
as
sides
of
consequents, the polygons
panying
Thus, in the accomproportional.
=
are
in
order,are
order,the polygons
pairsof equal
in
as
b : m
d :
c : n
figure,if a : I
polygonshave their sides proportional. The
ratio
4, find .4C.
=
polygon,taken
one
polygons,taken
two polygons.
the
=
POLYGONS
antecedents,form a series of
another
polygon,taken in order
are
Di?
30, ^li"
||.BC, AB=
SIMILAR
417.
28,
=
Prove.
20, find AE.
662.
Ex.
50,
=
If D^
661.
and.4C
AB
12, is Z"#
=
10, and
=
right,if AB
8, prove
at the
diagram
15, AC
12, AD
D," parallelto BC.
=
GEOMETRY
=
lines
lines is called
o
=
of the two
the
ratio
=
e
lines
:
p, the
forming any
polygons,and
of
two
similitude
the
of the
polygons.
419.
Def.
Two
and
equiangular
polygons
if their sides
are
are
if
they are
proportional.
similar
mutually
178
PLANE
422.
Cor.
Two
II.
423.
If
in.
Cor.
right triangles
is
angle of one
acute
GEOMETRY
equal
to
triangle, this line, with
a
which
triangle
Upon
663.
Ex.
other
to the
given line
a
parallel to
the
is similar
if
an
two
any
side
of
sides, forms
a
given triangle.
base
as
similar
angle of the other.
acute
an
line is drawn
a
are
construct
a
trianglesimilar
to
a
given triangle.
Draw
second
a
and
DF
thirds of AB.
two
altitudes of
Any two
inverselyas
other
each
to
to
the
lengths of
and
ABC
Compute
EF.
665.
Ex.
a
equal
DE
having
Estimate
triangleABC.
similar
triangleDEF
Draw
664.
Ex.
sides
the
triangleare
which
they
a
to
to
are
drawn.
tree, BC,
a
the
same
casts
What
^
arc
determine
to
height of the tree ?
the
is the
what
Ex.
that
CD
in
length
BC
found
What
To
A
BC
B.
to
was
then
was
and
line of
to
was
is found
the
to
distance
C
point was
perpendicular to
B,
a
90
to A
distance
across
AB
BC
EB
at
feet
at
C.
intersected
CE
measurement
feet and
across
located
off 100
measured
D
be 64 feet ;
?
perpendicular to
By
be
tree
find the
sightfrom
E.
at
CA
height of the
668.
river from
The
If
667.
Ex.
so
E
DF.
measurements
necessary
a
time
shadow,
a
a
pole, ED,
vertical
a
day
shadow,
casts
At
tain
cer-
a
of the
hour
CA.
At
666.
Ex.
210
was
feet.
the river ?
DF,
16
feet
its sides.
179
III
BOOK
669.
Two
isosceles
trianglesare
similar if the vertex
equals the
base angle
vertex
angle of
the
if
of the
other.
Ex.
follows
It
424.
" 419, and
from
other,or
similar
definition of
the
polygons,
that:
Prop. XIV
from
angle of
base
a
angle of one
one
equals a
are
equal.
(1) Homologous anglesof similar triangles
(2) Homologous sides of similar trianglesare proportional.
(3) Homologous sides of similar Mangles are the sides opposite
equal angles.
425.
In
Note.
therefore,it
case,
is desired
to
four
prove
lines pro-
triangleseach having two of the given
sides.
be
as
If, then, these trianglescan
proved similar,their
sides
will
be proportional. By marking xoith colored
homologous
crayon
be
the lines required in the proportion,the trianglescan
found.
readily
If it is desired to prove
the product of two
lines equal to the product of
the four lines proportionalby the method
two other lines,prove
just suggested,
portional,try
to
find
of
pair
a
lines
then
put
the
product
of the
length
of
equal to
extremes
the
product
of
the
means*
426.
a
The
Def.
circle is the
point and
the
length
second
from
secant
a
of the
included
segment
pointof
external
an
intersection
pointto
the
between
of the
and
secant
the circumference.
Ex.
the
product of the
the means,
Ex.
what
still hold
the
and
671.
are
Ex.
the
with
672.
point
of
secant
becomes
secant
which
Ex.
state
If two
intersect
Rotate
the
of
tangent.
product of
\p discussed
product
from
secants?
any
given point
the
Does
a
the
What
two
are
tangent
?
of Ex.
two
theorem
until
the
Does
671 about
the
rotating
segments
lines is meant
IV.
the
of the
the theorem
Prove.
aerain in Book
of
theorem.
of the secants
one
has become
a
the
chords.
circle,
to
a
of
Ex.
670
them?
to
intersection
a
as
drawn
are
of the
regard
tablish
circle,es-
a
equal to
result
secants
within
the segments of the
extremes
your
segments
670 stillhold ?
*By
wiH
chords
proportionalityamong
a
Place
If two
670.
product
of their
of
"n.
fjg^88
fC
f\
\
|
\
J
^\_ *S
\
measure-numbers.
This
180
PLANE
GEOMETRY
Proposition
Two
427.
XV.
triangles which
Theorem
have
their sides
proportional
similar.
are
E
A
Given
and
BEF
such
ABC
that
=
"
AB
To
A
prove
DEF
A
~
=
"
AG
BO
ABC.
Argument
1. On
lay off
BE
DH
Reasons
AB, and
"
on
DF
lay
1.
"54,14.
2.
"54,15.
3.
By hyp.
4.
"309.
5.
"415.
6.
"423.
7.
"424,2.
8.
By hyp.
9.
"391.
off BK=AC.
2.
Draw
'
""
BE
o
_BFf
AB~
AC
BE
.
4.
BF
.".
=
5.
6.
.\
.-.
A
BK
IIEF.
HK
DEF
It remains
A
~
BE
But
11.
Ex.
BHK
=
A
^"(7.
BH~
HK
BE
EF
BE
EF
BH
BC
I.e.
BC
.'.HK=BC.
Now
.\
12. But
13.
A
_EF
AB
10.
BHK.
to prove
"
9.
"
"
BH
.-.
A
BH=
Arg.
ABC.
11.
"116.
A
DiMr.
12.
Arg.
A
^BC.
13.
"309.
BHK=A
A
A
BEF~
BEF
~
If the
673.
respectively,
are
the
and
10.
AB
sides
DK=AC
Q.E.D.
of two
trianglesare 9, 12, 15, and
Explain.
trianglessimilar ?
1.
6.
6, 8, 10,
BOOK
674.
Ex.
shall be similar
to
diagonals.
a
a
Construct
675.
Ex.
trianglethat
given triangle.
Construct
a
do
How
Hint.
If
XVI.
the
Given
A
and
DEF
C
A
a
given perimeter and
Theorem
angle of
an
the
including
Ex.
to
one
prove
A
676.
of these
AB
Two
~
with
sides
corKe'|ponding
median
in
one
to
tional,
propor-
Z
A
=
Z
D
A
and
"
=
BF
DEF.
trianglesare
sides
equal
one
DE
To
?
similar.
trianglesare
ABO
have
other, and
tlie
angle of
an
triangles
two
shall have
trapezoid,given the two bases and the two
the diagonalsof a trapezoiddivide each other
Proposition
428.
181
III
similar if two
triangleare
in the other
sides and
proportionalto
triangle.
the median
two
drawn
sides and
the
182
GEOMETRY
PLANE
B
a
E
Fig.
Ex.
677.
Prove
In
Ex.
such
In
.
If in
679.
that
Ex.
AG
GA
AB
:
680.
triangleABC,
AB
=
:
Construct
of two
sum
429.
:
AD.
AC
=
each,
_L
To
^1Z?
.4Z",prove
a
XVII.
triangles that
perpendicular
or
BD
=
BC.
lar
G simi-
=
to
a
prove
to
their
sides
each,
are
parallel each
similar.
and
ABC
A
AB
1. AB, BC, and
C
~
CA
are
A'B\ B'c',and
A A, B, and
are
point
given triangleand having
A!B'c',with AS, BC, and
and
to a'b',b'c',
c'a'.
(Fig.2) respectively
to
a
BD.
A
sup.
II(Fig.1)
Reasons
IIor
_L
respectively
1.
By hyp.
2.
""198,201.
C'A'.
or
equal respectively
and c\
to A A',B ',
respectively
C are
CA
A'b'c'.
Argument
.*.
if 7) is
and
Theorem
have
each
BC,
=
1.
A
Given
2.
and
triangleAB
Prove
Fig.3, GA
trianglesimilar
equal to a given line.
sides
Two
Fig.
or
AE
=
Proposition
to
Fig.\,AB
triangleDBG.
to
Fig. 2, AB
DBC,
triangleADE.
to
the
similar
triangleABC
Ex.678.
C and
AB
triangles
2.
183
III
BOOK
Argument
3.
Reasons
follows
as
(1) Za
2 rt.
Z
(3)
C
A
A, Zc
Z
+
A of the two
But
6.
.-.
this is
(3)is the
i.e. the
7.
A
Zc'
=
two
AABC~AA'B'c'.
430.
Question.
SUMMARY
the
OP
I. Two
431.
"161,
4.
"54,2.
5.
"204.
6.
161,b.
a.
+
Zb'
=
2 rt. A.
2i"A,
=
Zb';
is more
hence,
of the
sum
than four rt. A.
impossible.
only suppositionadmissible;
.-.
and
3.
2 rt. A.
=
A', Zb
Z
=
c"
+
Z ".
also,Za=
According to (1)and (2)the
5.
fore,
there-
:
(2) Za=Za',Zb
Z
made,
Za'=2tLA,Zb+Zb'
+
=
4.
be
suppositions
may
Three
A
mutually equiangular.
q.e.d.
Can
other
are
two
one
pair
angles in Prop.
of
"420.
7.
XVII
he
mentary
supple-
pairsequal ?
CONDITIONS
FOR
trianglesare
SIMILARITY
similar
if
OF
TRIANGLES
angular.
they are mutually equi-
(a) Two
trianglesare similar if two angles of one are equal
to two angles of the other.
respectively
(6) Two right trianglesare similar if an acute angle of one
is equal to an acute angle of the other.
(c) If a line is drawn
parallelto any side of a triangle,
this line,with the other two
sides,forms a trianglewhich is
similar
II.
III.
equal
IV.
to
to the
Two
trianglesare
Two
to
given triangle.
an
Two
ea^h,or
similar
if their sides
are
proportional.
trianglesare similar if they have an angle of
angle of the other, and the including sides
trianglesare similar
perpendiculareach to
if their sides
each.
are
one
portional
pro-
each
parallel
184
PLANE
Ex.
circle
similar
a
Ex.
a
from
in
AQ
second
trianglesimilar to the
BC
arc
x
lines
ABC
P,
to
so
circle and
a
any
is
a
joining the mid-points of
the
inscribe
in
the
point of BC,
that
and
the
a
chord
in
circle.
a
is drawn
a
triangle
line is drawn
A
from
equals angle ABC.
angle ABQ
sides of
given triangle.
triangleinscribed
Prove
B
to
AB
point Q
a
x
AC
=
AB.
Proposition
The
432.
bisector
opposite side
other two
Given
To
about
triangle.
The
A
circumscribe
triangle.
triangleabout
a
683.
684.
Ex.
circle and
a
inscribed
to the
Circumscribe
682.
circle
trianglein
a
trianglesimilar
a
Ex.
form
Inscribe
681.
GEOMETRY
prove
into
of
an
segments
XVIII.
Theorem
angle of
a
which
triangle
are
proportional to
sides.
A
ABC
with
AP:PC=
BP
the bisector
AB:BC.
of Z
divides
ABC.
the
the
186
PLANE
434.
A
Def.
GEOMETRY
line is divided
internallyand externallyinto segments
callyequal ; thus, if line AC is divided
at Q
is
to PC
AP
that the ratio of
so
to QC, AC
The
bisectors
689.
Ex.
of
vertex
triangledivide
a
Divide
690.
Ex.
3 to 6 ; in the
ratio
a
6,
to
a
Proposition
In
435.
have
the
altitudes
AH
XX.
ratio
same
and
and
as
and
ABC
Q
harmonically.
exterior
angles
at
b
in
the
ratio
homologous
two
any
two
DEFf
with
homologous
two
tudes
alti-
sides.
corresponding
DK.
DK
Reasons
.-.
and
ABH
A
AABH~
DEK,
4.
'
oppositeZ B
DK, oppositeZ.E
=
AE.
AB,
DK~
DE
"422.
3.
" 424, 2.
4.
"424,2.
5.
"54,1.
FD
EF
=AB
2.
DE,o])ipositeZ.EKD
'
AH
"424,1.
oppositeZ.BHA
_
_
DE
1.
CA
BC
But^
B
DEK.
AH,
,
'
Z
CA
BC
_
=
EF
FD
of
Theorem
prove
In rt. A
any
given straightlines.
are
AH
To
ternally
ex-
harmonically.
triangles any
similar A
two
Given
similar
two
side
and
and
\
divided
be
interior
where
P
f
to
opposite
numeric
are
PC
given straightline harmonically
a
of
the
the
ratios
at
internally
|
is said
of
whose
^
numericallyequal to
the ratio of AQ
if it is divided
harmonically
Q.E.D.
book
Proposition
436.
If three
XXI.
point intersect two
segments of the parallelsare
common
Given
lines PA, PB, PC,
PD
the IIlines AB
intersecting
A\ B',C',d',respectively.
and
Ex.
691.
If three
segments
on
or
two
Theorem
straight lines
more
or
187
in
drawn
through
a
parallels, the corresponding
proportional.
drawn
and
through
A'd' at
a
common
point P
pointsA, B, C, D
and
portional
non-parallelstraightlines intercept proa
common
point.
parallels,
they pass through
more
188
Ex.
A
692.
miles
a
hour
an
is
man
on
walk
at
Ex.
in the
sides of
third side is 6.
12.
to
Compute
In
two
ratio
as
694.
Ex.
in the
same
in the
ratio
same
437.
Make
the
the
rate
does
8 and
the
between
4 feet from
he
walk
the
the
pedestrian, does
side
the
footpath
pedestrian
?
11, and the altitude upon
side homologous to 8 equal
triangleas you can.
two
homologous
triangles,
any
homologous sides.
medians
are
two
to
Scale.
on
paper
in which
of
your
top of your
the
Measure
each
line is
Check
desk.
-^
as
desk.
long
work
your
as
by
your drawing, and the corresponding
is called drawing to scale.
ing
Map draw-
diagonal of
the
principle.The scale of
represented: (1)by saying, Scale,-j^"or
inches
drawing the scale
; (2)by actually
be
drawing may
Scale,1 inch to
as
distance
other
are
line of your desk.
This
illustration
is a common
"
tree
the
bisectors
any
correspondingline
measuring
the
footpath on
of 30
rate
two
triangles,
homologous
any
homologous sides.
similar
drawing
a
a
trianglehas
two
any
Drawing
from
triangleare
a
similar
as
a
If the
parts of the second
many
In two
695.
Ex.
as
at the uniform
on
is 44 feet,and
similar
A
while
If so, at what
?
rate
Two
693.
road,
a
chauffeur
the
automobile
an
opposite direction.
track
auto
uniform
a
side of
one
walking
riding in
is
man
footpath and the
continually hides
the
GEOMETRY
PLANE
of this
"
"
12
indicated.
B
; 3 feet 3 inches.
24 inches
On
697.
Ex.
draw,
above,
is 28
the
paper
to
represent
B
scale
the
ter
diame-
The
698.
a
scale
the
costs
farm
figure
drawn
indicated.
of
cost
around
on
feet.
represents
the
to
lines
board
black-
the
circle whose
a
Ex.
scale above, draw
the
Using
696.
Ex.
36
24
12
putting
farm, if
$2.50
to
Find
a
fence
the
ing
fenc-
per
rod.
40
160
III
BOOK
Proposition
XXII.
438.
If two polygons are
each
of triangles,similar
the
polygons are
189
Theorem
composed of the
to each
and
number
same
similarly placed,
similar.
C
J
E
polygons ABODE
Given
AACD
To
~
prove
A
FHI, A
ABE
polygon ABODE
and
~
A
FGHIJ
with
FIJ.
~
polygon FGHIJ.
A
ABC
~
A
FGHf
190
PLANE
439.
Any
Cor.
GEOMETRY
similar
two
into
the
and
similarly placed.
number
same
of triangles
Proposition
440.
Upon
line
a
to construct
XXIII.
homologous
polygon
a
polygons
similar
polygon
To
construct,
and
to
similar
MQ,
a
line MQ
a
to
side
of a given polygon,
the given polygon.
Draw
4.
At
homol.
polygon
I.
1.
to each
R
AD
on
each
divided
Problem
C
Given
be
may
~
to
side
AE.
polygon AD.
Construction
all
possiblediagonalsfrom A, as AC and AD.
2. At M, beginning with MQ
a
as
side,construct A 7, 8, and
9 equal respectively
to A 1,2, and 3.
" 125.
3. At Q, with MQ as a side,construct
Z 10 equal to A 4, and
prolong side QP until it meets ML at P.
" 125.
P, with
prolong side
5.
At
MP
OM
a
side,construct
as
a
at 0.
MR
side,construct
until it meets
ON
is the
as
until it meets
PO
0, with
prolong side
6.
PM
Zll
Z12
equal
ADE
~
A
ABC
~
.-.
polygon
2.
A6,
" 125.
Proof
Argument
1. A
to
polygon required.
II.
A
MPQ,
A
MNO.
MP
~
A
Reasons
ACD
~
A
polygon AD.
MOP,
and
" 125.
at N.
MF
equal to Z5,
and
q.e.d.
1.
"421.
2.
" 438.
and
discussion is left
The
III.
other
Theorem
homologous
two
any
sides a,
~
homol.
perimeter
_
of P
*"
prove
n, o,
b,
^
k
I
a
2.
f
-"
a
of Q
k
Reasons
+ b+
c
I +m
/
"
o
nIV
mIIV
k +
/
-.
!L"
U
e, and
andjp.
Argument
IV
d,
c,
=
"
perimeter
a_^_
to each
polygons are
sides.
polygons P and Q, with
to sides k, I,m,
respectively
Given
To
exercise for the student.
an
perimeters of two similar
The
as
as
XXIV.
Proposition
441.
191
in
book
?i
"419.
2.
" 401.
3.
" 309.
JJ
+d+e+f=a
+
1.
p
\J
+ o-|-/) "
ThatiS,PerimeterofP
"
=
perimeter of
Ex.
a
as
Find
the
Ex.
701.
The
any
two
Ex.
702.
perimeters of
homologous medians.
triangleat its
Q.E.D.
k
two
similar
homologous
perimeters of two
homologous diagonals.
The
two
perimeters of
first is 8.
700.
Ex.
any
The
699.
side of the
Q
two
similar
similar
If
perpendiculars are drawn
extremities,and if the other
prolonged
to meet
fivf,
triangleseach
these
perpendiculars,the
of which
#issimilar
to
any
polygons
polygons
two
to
are
138 ;
each
other
to each
as
other
hypotenuse of a right
sides of the triangleare
the
figurethus
one
and
second.
trianglesare
to
152
are
side of the
formed
of the others.
contains
192
PLANE
GEOMETRY
Proposition
In
442.
The
and
The
II.
hypotenuse
and
to that
rt. A
I. A
formed
are
similar
to the
given
a
proportional between
mean
the
a
the
proportional between
segment of the hypotenuse
mean
the
side.
and
ABC
BCD
the altitude
BD:
CD=
AB
:
ABC, A
A
~
CD
CD
upon
the
hypotenuse.
ADC
A
~
ABC, and A
BCD
~
A
ADC.
DA.
:
and
BC=BC:DB
I.
AC=
AB:
AC:
AD.
Argument
1.
In rt. A
2.
.'.
3.
In rt. A
4.
.-.
5.
.:Z1
6.
.'.
A
A
A
II.
Hint.
the
the
:
prove
III.
is
is
adjacent
II.
upon
other.
side
Given
altitude
of the hypotenuse.
Either
whole
To
to each
altitude
segments
III.
the
is drawn:
triangles thus
triangle
Theorem
right triangle, if
a
hypotenuse
I.
XXV.
BCD
A
~
~
=
/.B.
ABC,
A
ABC.
A
ADC.
Za
=
A A.
Zb.
=
BCD
~
proof of
Mark
proportion.
the fact that
ABC, Z.B
ABC.
and
ADC
ADC
The
and
B CD
Q.E.D.
II is left
(with colored
Decide
homologous
which
sides
as
an
exercise
for the student
chalk, if convenient) the lines required in
triangleswill furnish these lines,and use
of similar trianglesare proportional.
194
PLANE
GEOMETRY
XXVI.
Proposition
To
445.
construct
a
Problem
proportional between
mean
two
given lines.
t'
/
f
I
!0
rr^rr^
3
To
lines,m
two
Given
construct
and
line Z
a
N
w.
that m:l
so
=
l:n.
I. Construction
1.
CZ)
=
2.
On
With
At
lay
AB,
off
AG
and
m
=
mid-pointof
0, the
describe
construct
G
AD,
as
and
center
with
a
radius
semicircumference.
a
GEWAD,
the semicircumference
meeting
" 148.
at E.
4.
line,as
n.
equal to OB,
3.
indefinite str.
any
is the
CE
II.
The
required line
proof and
I.
discussion
left
are
as
exercise for the
an
student.
Ex.
706.
By
lines
given
two
of " 444, II, construct
means
by
method
a
different
a
tween
proportional be-
mean
from
that
given
in Prop.
XXVI.
Ex.
V3 ab,
to
Use
707.
a
and
b
the method
of
being given
lines.
Let
Analysis.
Then
.-.
,\
to
3
Ex.
708.
Construct
Ex.
709.
Using
VS
;
V5
a/2,2\%
;
V6.
5\/5.
a
a
line
3
a
line
one
Choosing
Prop.
x
=
x*
=
3 ab.
=
x
:
equal to
your
a
line
equal
required line.
V3ab.
=
inch
to construct
the
x
: x
XXVI
b.
a
long
own
V3,
where
as
unit,construct
a
a
unit, construct
is
a
a
given line.
a
line
line
equal
equal
to
book
XXVII.
Proposition
In
443.
is
equal
447.
square
448.
is
to the
Cor.
is
of
of the
sum
The
I.
Cor.
to
root
The
its side
of
1.
2.
in
Prop.
2000
b.c,
of
d
two
sides.
of a right
hypotenuse minus
angle
trithe
square
the
+
Proof
s2=2s2.
s-y/2.
=
Note.
was
having made
cor* measured
of the
side
"
two.
Historical
XXVII
of the othe?
multiplied by
d2=s2
.\
of the hypotenuse
square
either
diagonal of a
Outline
449.
of
square
to
n.
Theorem
squares
the square
the other side.
equal
equal
square
right triangle the
any
195
in
off into
known
q.e.d.
The
property of
the
right triangle stated
early date, the ancient Egyptians,
three pegs a
a right triangleby stretching around
See Note, Book
3, 4, and 5 units.
IV, " 510.
at
a
very
196
PLANE
Ex.
710.
By
Ex.
711.
If
Ex.
712.
The
Find
other
the
of
side
a
of
and
arm
the
the
713.
Find
the
714.
Find
a
715.
Divide
its
of
diagonal.
15 and
one
the
equilateraltrianglewhose
an
is 9.
arm
hypotenuse made
rightangle.
of the
altitude of
equal to V2 inches.
line
righttriangleis
a
segments
vertex
a
inches, find
is 6
square
hypotenuse
perpendicular from
Ex.
of " 448, construct
means
a
GEOMETRY
the
by
side is 6
inches.
Ex.
side
of
altitude
equilateraltriangle whose
an
is
8 inches.
Ex.
Ex.
The
716.
chord
6 inches
Ex.
a
Ex.
of
the
center
The
chord
'whose
718.
Tn
a
the center
The
which
shall be
circle is 10 inches.
a
; 4 inches
chord
is
chord
a
in the
ratio
Thus,
C
of the second
upon
line
a
line included
projectionof
and
MN}
is the
CD
far from
the
length is 28 inches
long is 5 inches from
of the first line upon
is the
How
12 inches
A
upon
?
foot
a line is
upon
the projections
segment
between
the second.
D
MN,
of
projection
AB
is the
upon
projection
MN.
9
B
B
ter
cen-
?
the center.
a
of
a
center.
length is
whose
length of
the
point upon a line is the
the point to the line.
projection
of the extremities
Find
the
? whose
24 inches
projection of
The
Def.
from
circle is 20 inches.
a
perpendicularfrom
the segment
B
segments
length is 32 inches
circle
a
Def.
451.
of
of
radius
717.
450.
of the
radius
from
far from
How
into
V2.
of 1 to
is
line
a
A.
Irk
M~C
D
C
M
N
*
V
M
A
Ex.
CD
be
ever
a
In the
719.
of
AB
be
on
M2f
equal
to
AB
a
maximum
?
?
than
greater
a
condition
what
minimum
AB
?
?
Will
Will
the
will the
tion
projec-
the
projection
projectionever
point ?
Ex.
Ex.
721.
another
right isosceles trianglethe hypotenuse of which is 10
length of the projectionof either arm upon the hypotenuse.
In
720.
inches,find the
upon
figuresabove, under
be
a
Find
if each
the
projectionof
side is 6 inches.
one
side of
an
equilateraltriangle
BOOK
722.
Ex.
of the other
each
in
(3)
each
in
an
(1)
Draw
triangle.
:
side of
the shortest
angle is 60".
longer if the
Ex.
of
Find
triangleupon
a righttriangle;
triangle; (2)
of
the
projections
longest side in
the
projectionof
the
upon
the shorter
the
their
longer.
side upon
the
; 45".
Parallel lines that have
725.
and
inches
12
projectionof the shorter
angle is 30"
included
and
8
triangle are
a
the
723, find
Ex.
In
724.
Ex.
sides
Two
723.
included
equal projectionson
the
same
line
equal.
XXVIII.
Proposition
In
452.
the side
sides and
triangle to find
any
the value
of the square
of the other
upon
of
two
the
A
X
Fig.
D
1.
with
ABAX,
Given
Z.X
acute; and
the
p,
projectionof
b
a.
upon
To
Problem
opposite an acute angle in terms
of the projectionof either of these sides
other.
of a2 in terms
the value
find
of a,
b, and
p.
Argument
1.
In
2.
But
3.
And
4.
a
in
acute
case.
Ex.
are
sides
obtuse
an
projectionsof
the
Draw
197
III
rt. A
AD2
DB
.-.
5.
DB*
.-.
453.
=
=
=
b2
a2
=
+
DB2.
(Fig.1)or
"
"
2 ap
p2 -f
a2 + b2
"
Why
Question.
for
aad^discussion
AD2
=
p2.
-
a"p
a,*2
b2
=
i.e. x?
a?
BAD,
Reasons
a
+
a2
p
"
a
(Fig.2)
p2.
"
2 ap -f
2 ap.
is it not
right triangle ?
p2 ;
1.
"446.
2.
"447.
3.
" 54, 11.
4.
" 54, 13.
5.
"309.
.
q.e.f.
necessary
to
include
here
the
figure
198
PLANE
454.
as
In any
is
XXVIII
Prop.
theorem
follows
to
the
twice
by
the
of the other
726.
Ex.
the squares
product of
side upon
If the
727.
sides of
of
arm
b and
728.
729.
730.
Ex.
Find
x
form
oppositean
of
the other
of
a
of
these
angle
acute^
ished
dimin-
sides
two
sides and
?
Why
of
7, 8,
and
the
10,
tion
projec-
angle
is the
?
Prop.
XXVIII
to
(in the figurefor Prop. XXVIII)
of
the square
an
upon
a
If two
a
the
sides of
a
In
any
the obtuse
the other
two
obtuse
angle
sides
these sides and
Given
A
BAX
is
with
triangleare
(c2=
4 and
XXIX.
equal
a
and
12 and
the
jection
pro-
to the
obtuse,and
a2 + b2 + 2 ap
projectionof
triangle
of the side
square
sum
by twice
projectionof the
Z.X
the
Theorem
increased
a.
prove
of
second.
triangle,the
the
15, find
13, 14, and
triangleare
is 2, find the third side of the
the second
Proposition
455.
in terms
b.
If the sides of
the first side upon
To
side
triangleare
a
acute
of the first side upon
upon
one
the statement
projectionof
the
Ex.
of
the
righttriangle.
a
Ex.
Apply
in
it.
opposite 10 obtuse,right,or
Ex.
of the
square
of
sum
stated
be
may
:
the
triangle,
equal
GEOMETRY
of
the
site
oppo-
the squares
product of
other side upon
p, the
of
one
it.
projectionof
b
BOOK
199
111
Argument
1. In
rt. A
2.
But
3.
And
4.
DB
BB2
5
.-.
a?=Al? + Bb\
b2-p\
1.
"446.
2.
"447.
=
a+p.
3.
"54,11.
=
a2 +
4.
" 54, 13.
=
62-p2+a2
5.
"309.
BAD,
AB2
.\
^
i.e.
x1
=
+p\
2 ap
2ap+P2;
+
a2 + b2 -f 2 ap.
=
q.e.d.
and
Props. XXVIII
From
456.
Reasons
following formulas
of a triangleupon
sides of a triangle:
the
computing
for
Prop. XXVIII,
p
b*
+
we
derive
may
projectionof
if a,
; thus
another
^
From
XXIX,
"~
=
6, and
the
side
one
represent the
c
^
(1)
g
(2)
2a
From
It is
are
a* +
Prop. XXIX,
seen
that
identical
p
-
the second
and
that
"2
~
=
of these
members
equations
only in sign.
computing the
differ
first members
the
two
Hence, formula
(1) may always be used for
length of a projection. It need only be remembered
that if p
is
positivein
is
negative,the
obtuse.
It can
likewise
be shown
(seeProp. XXVII)
0, the angle oppositec is a rightangle.
p
c
if
is acute;
while
if p
that the
angle opposite
angle oppositec
indicates
is
that
=
Ex.
731.
Write
Ex.
732.
In
of b upon
Ex.
Is
it
calculation,
any
733.
angle
Ex.
735.
736.
projectionof
In
the
=
obtuse
sides of
The
20
sides of
a
If two
the
sides of
a
third side.
the
a
upon
25;
=
b.
find the
tion
projec-
?
projectionof
a
upon
b.
?
obtuse
triangleare
of the
is the value
c
obtuse
732, find
triangleare
a
20,
=
Ex.
acute, right,or
If two
angle is 45",find
triangleof
right,or
angle is 120", what
Ex.
for the
c.
opposite the side
Ex.
formula
15, 6
triangleABC, a
Is angle A acute, right,or
C acute,
734.
the
8, 14,
and
20.
Is the
angle
?
10 and
12, and
their included
16, and
their included
third side ?
triangleare
12 and
200
Ex.
If in
737.
PLANE
GEOMETRY
triangleABC,
angle C
AS2
If
738.
Ex.
ABC,
BC2
=
line is drawn
a
base
meeting
AB
AC2
+
Proposition
In
457.
equal to twice
increased
by twice the
sides
is
at
AB.
=
the
C of
prove
an
isosceles
that
of the
square
of
the
of half
the
of
squares
median
any
A
To
with
ABC
b2 +
prove
c2
ma,
2
=
the median
(-Y
+ 2
to side
side
upon
that
a.
ma\
Reasons
Argument
1.
Suppose
and
Z
6 "
2.
Let p be the
3.
Then
4.
And
from
"
A D
.-.
is obtuse
ma
upon
a.
C,
ABD,
(2) ma2
+
"
5.
Z ADC
of
projection
A
A
; then
1. " 173.
is acute.
BDA
from
c
"""
2.
"451.
3.
"455.
4.
"454.
p.
62-hc2=2^J+2ma2.
Q.E.D.
5.
two
third
side.
Given
angle
tri-
Theorem
sum
square
BC.
vertex
D,
that
BD.
XXX.
triangle,the
any
the
prolonged
dl"2-CBi
AC-
+
from
120",prove
=
"54,2.
202
GEOMETRY
PLANE
Proposition
XXXI.
If through a point
the product of the
460.
drawn,
chords
is
to the
equal
Theorem
within
two
chords
circle two
a
of
segments
product of the
two
circle 0, and
AB
are
of these
one
of the
segments
other.
Given
P,
chords
To
drawn
6,
=
In
751.
find
and
left
as
"
A
~
two
CD, any
PD.
exercise
an
and
for the student.
PDB.
the
figure for Prop. XXXI,
the
same
if PA
5,
=
PB
12, and
=
PC.
In
752.
Ex.
PC
=
A APC
Prove
Ex.
PB
"
proof is
Hint.
PA
PA
prove
The
PD
point within
through P.
a
figure,if PC
10, PD
=
8, and AB
=
21, find
=
PB.
find AP
the
In
753.
Ex.
and
figure,if PC
same
6, DC
=
=
22, and
AB
=
20,
PB.
the
Ex.
754.
In
Ex.
755.
If two
figure,if PA
same
PC
m,
=
n,
=
and
PD
"
r, find
PB.
and
10, and
second
second
a
circle
are
the
of
are
lengths 8
segments
of the
?
Ex.
756.
between
Ex.
one
the
intersectingwithin
bisects the first,what
chords
two
Hint.
If two
757.
chord
construct
of Prop.
By means
lines.
given
are
the
a
and
chords
b
segments
Find
the
locus
construct
intersect within
inches,
of the
XXXI
while
second
of the
the
second
a
a
circle and
chord
proportional
mean
the
segments of
d
measures
chord.
mid-points of chords
equal
to d.
inches,
CE
=
If two
758.
Ex.
then
ED,
"
a
and
lines AB
203
III
BOOK
intersect
CD
circumference
be
can
E
at
that
so
the
passed through
AE
four
"
EB
points
A, B, C, D.
XXXII.
Proposition
Theorem
// a tangent and a secant are
given point to a circle, the tangent
drawn
461.
the whole
between
tangent
Given
circle
the
to
and
PT
and
secant
secant
is
mean
a
from
any
tional
propor-
its external
drawn
PB
from
segment.
the
point P
0.
PB
PT
m
To
prove
"
-
:
PC
PT
Reasons
Argument
1.
Draw
2.
In
3.
Z2
4.
.-.
crand
A
PBT
BT.
and
.
CTP, Zp
Z2'.
=
APBT~
A
CTP.
oppositeZ
PT, oppositeZ
PT, oppositeZ
PC, oppositeZ
PB,
5
.
'
_
~
462.
Zp.
=
Cor.
I.
If
3 in A
PBT
3' in A
CTP
2 in A
2' in A
a
segment.
"54,15.
2.
Byiden.
3.
"
4.
"421.
5.
"424,2.
362, a.
PBT
Q.E.D.
CTP
tangent and
given point to a circle, the
equal to the product of the whole
any
1.
a
secant
square
secant
are
drawn
from
of the tangent
and
is
its external
204
PLANE
463.
If
II
Cor.
to
given point
any
its external
two
a
GEOMETRY
or
secants
more
drawn
are
circle,the product of any
from
and
secant
is constant.
segment
T^r'
secants
Given
drawn
from
be
denoted
by
respectively.
To
O, P
to circle
point P
let their external
and
PF,
PE,
PD,
segments
PA,
PC,
PB,
prove
PD
PA
"
PE
=
PB
"
=PF
"
PC,
Argument
1.
From
P
draw
to
tangent
a
circle 0,
as
PT.
PT2
2.
PF
3.
"
Ex.
PE"PB
=
If
and
Two
their external
their
common
762.
is the
mean
length of
to
to
two
If two
circumferences
:
ratio at
PB.
This
from
the
point to
same
long
a
external
is the
the
If
first
?
their
intersect,
into two
if
point.
segment of
the internal
secant
outside
an
from
any
point in
chord
common
longed)
(pro-
tangents.
common
P
circle from
intersectingcircles
equal.
line is said to be
mean
a
second
the
(prolonged) are
A
drawn
9, and
chord
Def.
Q.E.D.
secant
AB
division
:
AP
divided
in
whole
line and
one
"
as
the
golden
mean
part is
the other
=
is known
and
extreme
parts such that
the
proportionalbetween
Thus, AB is divided in extreme
and
AP
12 and
are
if it is divided
ratio
a
PC.
drawn
are
tangents
bisects their
464.
PF-
=
inches,respectively,how
secants
The
761.
Ex.
=
PT2
PB;
"
?
segments
is 8, what
Ex.
18
of the secant
760.
secant
PE
tangent and
a
6
measure
segment
=
PC.
759.
Ex.
PT2
PA;
"
PA
PD-
.'.
circle
PD
=
section.
a
part.
XXXIII.
Proposition
divide
To
465.
Pi-oblem
internallyin
line
a
205
III
BOOK
and
extreme
mean
ratio.
""w
x
"?
I
I
I
I
I
"S
f
XV
I
/
'
i
B
line AB.
Given
find, in ABf
To
t
P
A
such
pointP
a
that AB:AP
AP:
=
PB.
I. Construction
1. From
B
draw
2. With
O
as
From
A
3.
4.
AB
With
and
^4 5
and
center
draw
at C and
_L
50
with
BO
as
describe
radius
through 6, cutting the
secant
a
" 148.
\AB.
"
a
ence
circumfer-
D.
A
as
and
center
with
AC
as
radius
draw
CP,
cutting
at P.
5.
AB
:
^IP
AP
=
:
PP.
II.
Proof
Reasons
Argument
1.
AB
2.
.-.
AD
3.
.-.
AD
4.
.-.
AD"
5.
.'.
AP
6.
.'.
^1P:
HJ.
is
tangent
:
AB
=
^1P:
"
CD
:
The
cumferen
cir-
:
1.
"314.
2.
"461.
AC.
3.
" 399.
AP.
4! " 309.
to circle O.
AB
:
AB
=
AB
AB
=
AB"
AB
=
PB
AP
=
JP:
:
AC.
"
AC:
AP
:
^P.
PB.
discussion
Q.E.D.
is left
as
an
exercise
5.
" 309.
6.
" 395.
for the student.
206
PLANE
Ex.
Hint.
FA
In
Ex.
Then
if
is the
s
Hint.
Ex.
I
: s
:
A
I
P'^l
:
value
the
inches
10
767.
figure
A
to B
CD
=
be
125
Ex.
be
can
lengths of
line 8
inches
the two
the
accompanying
used
find
the
distance
to
of
by
CE
is the distance
AB
to obtain
feet.
He
height of
the
this method
Use
length of the
his
Was
tree.
for
and
tical
ver-
tree's
?
correct
answer
shadow
and
Draw
height of your
the
measuring
a
up
exactly6
quickly the
extreme
the
shadow
measured
and
?
the
then
extreme
to
until
pole measured
internallyin
%BC,
=
height of a tree in his yard. He set
pole 6 feet high and watched
of the
ratio,
from
measurement
little boy wished
A
768.
hill.
a
is found
What
s
mean
of I.
EXERCISES
the
feet.
in terms
longer segment.
how
\AC.
and
extreme
externallyin
Explain
ED
P', making
to
segments.
is divided
long
length of
the
oppositesides
on
of
is divided
long
MISCELLANEOUS
Ex.
ratio.
P'B.
:
internallyin
find
mean
BA
prolong
PA
=
and
extreme
s.
"
the
Find
ratio.
XXXIII
line I is divided
line
Find
766.
mean
s
=
A
ratio.
Ex.
externallyin
greater segment,
765.
mean
AB
prove
If the
764.
line AB
a
figure for Prop.
the
AD.
=
and
Divide
763.
GEOMETRY
called this the
figuresand
school
explain.
buildingand flag
pole.
If
Ex.
769.
AB,
is allowed
as
small aperture
and
W,
is formed
on
OE=
to
O, in
strike
inverted
an
light from
image
window
the
If the
=8
the
length of the tree AB=%5
th3
length of the
what
condition
This
Ex.
exercise
770.
between
Ex.
inches
two
771.
in
feet,find
Under
length of
illustrates the
By means
given lines.
In
a
length.
of
the
the
image equal
principleof
Prop.
certain circle
Find
tance
dis-
feet,and
CD.
image
will the
wall,
or
tree,as CD,
screen.
feet,OF
a
shutter
screen
of the
tree,
through
pass
white
a
30
a
a
a
XXXII
chord
length of
a
the
the
chord
the tree ?
photographer's camera.
construct
5V5
length of
inches
a
from
9 inches
proportional
mean
from
the
center
the
is 20
center.
internal
the
common
(2)
and
6, respectively,
what
is the
length of
774.
If from
Ex.
the
segments of
of
a2
2 ab
-
The
half their
than
+
that
of
Ex.
upon
Ex.
and
the
for
x.
c
it divides
and
b2
a2 + 2 ab +
unequal
two
lines is less
each
sides of
the
first from
If
783.
a
:
side
the
AB
AB"AB
3, what
feet.
the
are
If
line
a
lengths of the
?
figure,four
one
is 305
AB
of 2 to
BE.
:
lines whose
lengths shall
be
multipliedby V2, V3, 2, V5, respectively.
unit
altitude
the
BD
in the ratio
AC
Two
782.
length of
Prove
E.
Construct, in
given
from
triangle ABC
a
which
into
a
inches,
triangleinscribed in
the vertex
B, cutting Ay
isosceles
is drawn
divides
781.
Ex.
drawn
are
proportionalbetween
an
point,as
In
parallelto BC
segments
is
BD
in any
780.
Ex.
16
the tangent ?
?_5";
solve
c
proportionalbetween
mean
ABC
Chord
circle.
the base
secant
The
779.
Ex.
a
diagonals of a trapezoid divide
segments which are proportional.
into
other
and
sum,
778.
Ex.
is
righttriangle
long is
12, how
-
mean
a
8
b2.
+
777.
Ex.
a
Find
776.
tangent,
are
is 20.
centers
isosceles
an
X
Ex.
radii
whose
equationVOl+JIl
the
circles whose
tangent and
a
4 and
are
external
?
arm
point
a
secant
Given
775.
Ex.
each
two
between
hypotenuse
common
(1) the
:
to
tangent,
distance
the
If the
773.
Ex.
length of
the
Compute
772.
Ex.
207
III
BOOK
:
a
the
upon
b
=
c
triangleare 12 and
the opposite vertex
:
inches,and
is 9 inches.
the
dicular
perpen-
What
is the
side ?
second
d, show
18
that
K")!K6)=K")=K*"
Ex.
a
this fact into
translate
Also
If
784.
a
constant
proportion,will the
Ex.
li
785.
r
: s
=
a
verbal
statement.
is added
to
resultingnumbers
t
:
q, is3r
+
subtracted
or
be
in
from
proportion?
-:s=7t:2q?
Prove
each
term
Give
your
of
proof.
answer.
u
Ex.
units,what
Ex.
the
One
786.
the center
787.
of
a
segment of
a
circle is 4 units
is the
The
mid-points of
length of the
chord
drawn
through
long.
If the
diameter
other
segment
a
point 7
of the
units from
circle is 15
?
non-parallelsides of a trapezoid and the line joining
the parallelsides,if prolonged, are
concurrent.
208
PLANE
Ex.
Construct
788.
points and
be
of the two
790.
chord
Ex.
Ex.
tangent is a
from
circumference
Ex.
pass
to the
BE
a
circles
the
The
t2
with
ac
=
Then
sides is
third
In
equal
side
ac
the
of
an
angle of
Prove
the
greater
common
chords
a
t
:
dividing
r.
=
t+
=
t2 +
t2 =
triangle the
product
of the
diameter
of
ac
m
"
: c.
rs.
rs.
product of
altitude
the
upon
~AEBC.
A ABB
ac
=
Jut.
two
the
circumscribed
Then
prove
triangleis equal
by the product of the
Proof
P + tm
=
a
bisector.
circle.
Hint.
external
common
other,their
each
the
and
of
that
any
to the
and
that
proportionallyby the
divided
angle diminished
by
s
.*.
798.
are
rs.
"
3.
Ex.
Prove
the
their diameters.
t, the bisector of ZB,
segments
Prove
2.
CAB.
triangle
Find
circle.
of this
Outline
1.
opposite side is b.
other, their
each
of the bisector
square
A ABC
prove
"
point.
side b into the two
To
to the
of contact
segments of the third side made
Given
CD.
triangleis a, and the perpendicular
circles intersect
of the sides
product
:
R
ting
cut-
all chords of
tangent internally,
are
point
common
797.
CA
A^^%^\.
is any
AB,
arc
altitudes of
two
are
circles touch
If three
through
of
=
isosceles
an
of the smaller
796.
Ex.
CA
:
proportionalbetween
If two
circle drawn
bisector
BC.
:
mean
795.
lengths
^-
of
and
If two
794.
by
the
the
righttriangle,
ter
given its perime-
a
base
AD
CA
Ex.
point C
CE
Prove
CE
circle,and
a
extremity of the base
the equal sides.
AD:BE=
Ex.
of
an
793.
Compute
10, 12, 15.
least side is divided
the
angle.
The
792.
lengthsof
chord
a
Construct
let fall from
given
two
"
D.
acute
an
Ex.
is
at
791.
and
triangleare
a
through the middle
AB
through
pass
angle.
AB
drawn
chord
shall
which
given straightline.
a
into which
segments
of the opposite
Ex.
to
sides of
The
789.
Ex.
circle
a
tangent
GEOMETRY
210
GEOMETRY
PLANE
G
U
H
E
Fig.
times
without
which
is
a
2.
remainder
aliquotpart
an
Rectangle
EG
(seeFig. 2),then by taking a
of U,
as
one
fourth
Fig.
be
as
a
of U, and
square
applying
H
E
it
8 U+.
=
measure
obtained
3.
Rectangle
EG
=
*" U+=
U\ U+.
the
rectangle (seeFig. 3) a
which, divided by four,* will give
to
number
will
another
(and
G
Fig.
4.
Rectangle
AO
=
^U+=
111 U+.
usuallycloser)approximate area of the given rectangle. By
mations
proceedingin this way (seeFig. 4),closer and closer approxiof the
*
true
It takes
area
may
four of the small
be obtained.
squares
to make
the
unit
itself.
BOOK
aliquotpart of U,
an
integralnumber
the
If
470.
the
to
of
the
471.
be
closer
no
these
as
the
tions
approximawhich
square
of the
the
of
is
an
rectangle
ever,
is,howclosely
smaller
approach
unit
rectanglewhich
a
is the
square
measure-numbers
of the
subdivisions
the
measure-number
respect to
a
chosen
The
Def.
measure-numbers
473.
subdivisions
area,
chosen
approximate
as
or
brevitythe expression, the
472.
unit
the
of
unit
limit
tangle
rec-
approach
square
limit.
a
For
the
of
closer
aliquotpart
an
square,
with
successive
approach
Def.
of
of
area
figure,is used
a
to
figure with
the
surface
of
the
two
surfaces
is the ratio of their
unit.
ratio
of any
the
(based on
Equivalent
unit).
same
figures
figures which
are
have
the
area.
The
student
should
Equal figureshave
be made
that
note
the
:
shape and
same
figureshave the same
Equivalent figureshave the
Ex.
*
+
For
size; such
figurescan
to coincide.
Similar
(h
and
obtained, but
measure-number,
is incommensurable
as
an
rectangle also
given rectangle and
be also
will
The
Def.
which
is
limit.*
a
as
same
the
of
of the unit
subdivisions
mean
in the
unit
which
magnitudes). There
is approached more
definite limit which
and more
approximations obtained by using smaller and
a
zero
is contained
the
of times.
may
U
be found
may
of incommensurable
(by definition
zero
square
which
and
sides
area
aliquot part
by
a
of
given rectangle and
incommensurable, then
are
square
the
commensurable,
are
square
of
sides
the
If
469.
211
IV
Draw
799.
of these
none
\)(b + 1),
less
two
where
than
the
measure
measure,
size.
same
equivalent figuresthat
approximations
the
shape.
and
in
exceed
can
applied
a
is contained
the
base
b
are
certain
in
times
the
with
not
fixed
equal.
a
for
number,
altitude
h times
remainder
example
with
less
than
a
mainder
re-
the
212
PLANE
Ex.
a
equal figures on the blackboard
that equal figuresmay
and show
be added
to
paper,
that the resultingfiguresare not equal. Are
they the
of
out
way
Ex.
to
Draw
800.
GEOMETRY
801.
two
Draw
figuresto
equal figures,do not
Fig.
Ex.
802.
show
that
axioms
give results which
satisfythe
same
size ?
applied
equal figures.
test for
Fig.
1 above
in such
2, 7, and 8, when
1.
Fig.
them
cut
or
them
2.
card
containing 64 small squares,
cut into four pieces,I, II,III, and
IV.
Fig. 2 represents these four
pieces placed together in different positionsforming, as it would
a
seem,
rectangle containing 65 of these small squares.
By your knowledge of
similar triangles,
try to explain the fallacyin the construction.
474.
Historical
surveying,and the
Egypt and Babylon.
tablets,and
rectilinear
made
on
give methods
figures,and
changes
to devise
This
work
was
done
extant
is that
known
as
the
of
a
copy
of
an
for
older
records
states
in the
by
had
of
state
that it had
its
Babylon
made
were
circle.
that
The
the fact that
of taxable
amount
the
origin
beginning
on
clay
of
areas
Egyptian
its
several
records
were
inundations
land, rendered
it
of
sary
neces-
measurements.
the
Ahmes,
for
to have
supposed
finding the approximate
also of the
Rhind
It is called "Directions
be
The
land
accurate
is
earliest traditions
Herodotus
papyrus.
the Nile caused
a
Geometry
Note.
in land
in
represents
Egyptian priests,and
who
lived about
papyrus,
is
knowing
all dark
1700
preserved in
manuscript, dating
about
the earliest
the
script
manuscript,
manu-
British Museum.
things," and
3400
This
b.c.
b.c.
is
In
thought
addition
to
to
discussion
of areas.
Problems
a
on
problems in arithmetic it contains
which
show
some
pyramids follow,
knowledge of the properties of similar
and
which
and
of
figures
trigonometry,
give dimensions, agreeing closely
with those of the great pyramids of Egypt.
The
of the
geometry
Greeks,
of the
which
was
Egyptians
was
concrete
and
logicaland deductive,even
unlike
practical,
from
its
that
beginning.
BOOK
I.
Proposition
The
475.
Theorem
rectangle
a
its altitude.
and
its base
of
area
213
IV
is
equal
product of
to the
(See " 476.)
C
B
U
A
U
unit of
be the chosen
To
the
prove
I. If
AD
and
AB
that
(a) Suppose
AD
=
each
are
altitude AB, and
side is
u.
commensurable
in
with
and
AD
Argument
1.
Lay off u upon AD
Suppose that u
pointsof
At the
erect
3.
and
each
Js to
is contained
in
AD
r
times.
s
division
and
AD
respectively. 1. "335.
AB,
AD
on
AB,
and
A B
on
2.
" 63.
7.
" 309.
respectively.
is divided
rectangle ABCD
Then
AB
Reasons
in AB
times, and
2.
u.
of times.
number
into
unit squares.
4.
There
are
row
on
AD,
in
squares
5.
.-.
the
6.
But
of
area
r
of AD
of these
r
and
s
and
and
of
.-'.*"fche
area
s
squares
of
rows
in
a
these
rectangleABCD.
ABCD
=
r "s.
are
the
AB,
referred
respectively,
to the linear unit
7.
unit
A BCD
measure-numbers
u.
=
AD
"
AB.
let
AB.
"
is contained
u
and
AD
surface,whose
of ABCD
area
base
with
rectangleABCD,
Given
u
D
q.e.d.
an
tegral
in-
214
PLANE
(b) If
if
is not
u
The
of
measure
a
aliquotpart of
some
GEOMETRY
proof is left
is such
u
to the
and
AD
AB,
but
respectively,
measure.
a
student.
U
QD
A
If AD
II.
and
AB
each
are
u
ble with
incommensura
Reasons
Argument
1.
Let
be
m
to
measure
a
times
remainder,
PB,
as
2.
3.
draw
Now
Q draw
J_
PiW
J. Q
and
with
the
.*.
5.
Now
as
on
" 339.
2.
" 63.
3.
" 337.
A P.
4.
" 475, I.
No
5.
" 335.
6.
Arg. 5.
many
will
There
AD,
and
AB,
each
on
be
a
mainder,
re-
a
less
AB.
^LP
the
of
area
take
a
taken, when
and
to AD
each
are
measure
with
rable
commensu-
and
m,
smaller
small
Q-
of
measure
a
hence
u.
rectangleAPMQ=A
how
matter
^-D, and through
W_L
commensurable
4.
iB
1.
as
m
m.
Through
P
QD,
as
Apply
u.
and
AD
possible.
as
than
of
measure
a
u.
u.
of
measure
u
is
it is
appliedas a measure
AB, the remainders,QD
PP, will be smaller
than
the
measure
AQ
and
and
taken.
6.
.-.
the
may
less
difference between
be made
than
to become
any
segment, however
and
AD
remain
previouslyassigned
small.
215
IV
BOOK
Reasons
Argument
and
difference
the
Likewise
7.
AB
be made
may
less than
remain
between
AF
become
and
to
approaches AD
approaches AB as
Q
.-. A
9.
8.
" 349.
9.
" 477.
small.
limit,and
a
as
AP
limit.
a
approaches
AQ-AP
.-.
Arg.
signed
previously as-
any
segment, however
8.
5.
7.
AD-AB
a
as
limit.
10.
difference
the
Again,
and
A BCD
and
remain
between
be
may
made
less than
11.
assigned area, however
.-. APMQ
approaches ABCD
1 2.
But
the
to AQ
13.
.-.
the
of APMQ
area
10.
Arg.
5.
11.
" 349.
12.
Arg.
13.
" 355.
AB
rable
incommensu-
become
to
previously
any
.
APMQ
small.
as
is
a
limit.
always equal
4.
AP.
"
of
area
ABCD
AD-AB.
=
Q.E.D.
III.
If
AD
with
u,
476.
the
is commensurable
proof is
The
Note.
the
By
segment
477.
there
belongs a
If each of
finitelimit,not
the
478.
of
Cor.
I.
exercise
for the
lines is meant
two
The
but
u
an
proof
is
measure-number
finitenumber
any
that
given
of
the limit
of
in
to
the
every
of
product
straight line
" 595.
variables
their
student.
approaches
product
is
equal
a
to
(See " 593.)
limits.
The
as
of
lines.
zero, then
product of their
left
product
of the
measure-numbers
with
of
area
is
square
a
to the
equal
square
its side.
479.
Cor.
Any
n.
rectangles
two
the products of their bases
Outline
of
Then
b and
R
=
b
their
Denote
the
b',and
their
Proof.
R\ their bases by
and
"
h and
R'
=
to
are
each
other
as
altitudes.
rectanglesby R and
altitudes by h and h',respectively.
two
b '" h'.
R
Bf
b-h
b'
"
h''
216
PLANE
480.
Cor.
each
to
other
having
Two
(a)
III.
GEOMETRY
rectangleshaving equal bases
their
as
altitudes, and
equal altitudes
to each
are
Outline
^
(a\
Ex.
Ex.
Ex.
804.
show
how
Find
the
Find
805.
Ex.
area
If the
807.
area
of
area
Ex.
808.
Find
Ex.
809.
Find
"
j
"
"
_
is 7 units
whose
and
altitude
it contains.
squares
of
a
rectangle whose
base
of
a
rectangle whose
diagonal is
and
base
the
Ex.
of
d and
s,
Ex.
is 12
inches
and
10
inches,
feet,and
square
the
base,
square
two
Construct
a
2\ inches
rectangleare
whose
diagonal is
approximations
square
\\
rectangle.
roots
inches.
of
area
3
times
for
tenths
to
V2
8
the
to
V 5, respectively,
using
the
and
a
2 for
the
tangle
rec-
the
next, to
etc.
while
respectively,
811.
a
10 and
next,
Compare
810.
of the
area
successive
V
are
altitude of
the
area
the
for the
hundredths
rectangle is 60
a
approximation, taking
first
rectanglesif
a
diagonal and
a
a
rectanglewhose
a
rectanglewhich
diagonal and
side
of the
other
shall be
area
side
a
of
d' and
are
three
one
times
are
s'.
that
given rectangle.
a
Ex.
Construct
812.
ratio of two
Ex.
813.
bases
whose
Ex.
814.
thirds
Ex.
815.
6 inches
and
given lines,m
Compare
15
are
From
that
inches
a
of the
and
4
3
given
and
a
given rectanglein
n.
altitudes
are
equal,
but
inches, respectively.
given rectangle cut
If the base
and
shall be to
rectangles whose
two
and
and 8 inches, respectively,
are
base
fr
;?_
altitude ?
is the
if its sides
two
bases.
is 6 inches.
find
inches,"respectively,
the
unit
many
sides
If the
806.
inches,what
of
(1\
rectangle whose
the
of whose
one
Ex.
5
and
a
their
as
altitude is 5 inches.
whose
and
^L
"
Draw
other
rectangles
Proof
op
"
803.
is 4 units
^
'
"
(b) two
are
off
rectanglewhose
a
area
is
one.
altitude
the
base
of
a
and
certain
rectangle are
altitude of
inches,respectively,compare
their
a
second
areas.
12 inches
rectangle
218
PLANE
GEOMETRY
Proposition
485.
The
of its base
III.
Theorem
half the product
of a triangle equals one
area
and
its altitude.
X
B
A
A
Given
To
b
ABC,
with
area
of A
prove
C
altitude h.
b and
base
ABC
"b-h.
=.
Argument
1.
these
2.
line
II CB,
a
line
II CA.
lines intersect
at X.
Then
5.
/.
But
Cor.
altitudes
487.
|D
=
AXBC
b"h.
=
ABC
AXBC.
Jb
=
*
h.
Any
II.
the
their altitudes
""'
488.
to
each
two
by
A
by
to each
triangles are
two
their
and
Outline
f
equal bases
equal
and
equivalent.
products of their bases
Denote
"
Triangles having
I.
are
Cor.
O.
a
ABC
of A
.-. area
486.
A
of
area
a
draw
is
AXBC
3.
4.
draw
Through A
through B
T
op
and
as
the
b and
b',and
altitudes.
Proof
T',their
bases
h',respectively.Then
h and
other
by
T
\
b
-
h and
///.
1*
Cor.
(a)
III.
other
as
\V
.h'~ b' h'
"
triangles having equal
Two
their
having equal altitudes
altitudes,
are
to each
and
as
(b) two
their
bases.
bases
are
triangles
Outline
/
W
h
\
ft"
T'~b-h'
the
parallelogram having
a
Ex.
816.
Draw
Ex.
817.
Find
and
four
the
of
area
inches
Ex.
and
819.
Find
the
Find
the
b'-h
included
the
b
br
of
area
to
whose
the
angle
two
60".
whose
smaller
base.
same
sides 8 inches
Find
the
area
perimeters are
angles are
base
equilateraltrianglehaving
an
half of
one
altitude.
and
rhombuses
and
respectively,
16 inches,
t
base
same
the ratio of two
Find
818.
^b-h
parallelogramhaving
a
and
inches,respectively,
included angle is 45".
Ex.
T
equivalentparallelogramson
12
if the
,b""
K
;
triangle is equivalent
A
IV.
Cor.
Proof
of
b-h
T
219
IV
BOOK
a
side
24
30".
equal
inches.
to 6
Ex.
820.
area
of
an
three
or
more
whose
equilateraltriangle
altitude
is
8 inches.
a
Ex.
821.
Ex.
822.
Construct
Find
locus of the
the
given triangleand standing
Construct
823.
Ex.
one
one
of
825.
Ex.
Ex.
lines
through
base.
same
given triangleand
ing
hav-
triangleequivalentto a given triangleand
to two
equal,respectively,
given lines.
ing
hav-
Construct
a
one
trianglesequivalentto
a
Divide
826.
base.
ing
hav-
a
of its sides
same
given triangleand
Construct
,
two
the
on
of all
triangleequivalentto
a
given line.
triangleequivalentto
its angles equal to a given angle.
824.
Ex.
vertices
the
a
a
equal to
of its sides
on
equivalenttriangles
a
triangleinto
three
equivalenttrianglesby drawing
of its vertices.
'
Ex.
|
of
its
a
Construct
827.
a
triangleequivalentto "
of
a
given triangle;
given triangle.
Ex.
828.
Construct
Ex.
829.
The
a
of
area
triangleequivalentto
a
rhombus
is
equal
given square.
a
to
one
half
the
product
of
diagonals.
Ex.
drawn
Ex.
830.
to
point in a diagonalof a parallelogramlines are
two
oppositevertices,
pairsof equivalenttrianglesare formed.
831.
to
Ex.
the
If from
the
832.
and.9 inches,
any
Two
lines
Find
the
joiningthe mid-point of the diagonal of a quadrilateral
opposite vertices divide the figureinto two equivalentparts.
area
of
and
respectively,
a
triangleif
the
included
two
of its sides are
angle
is 60",
6.inches
220
GEOMETRY
PLANE
IV.
Proposition
derive
To
490.
formula for
a
the
of
area
a
triangle in
of its sides.
terms
C
A
Given
a
sides a, b, and
with
ABC,
derive
To
D
B
a
and
Problem
formula
for the
area
C
c.
of A
ABC
in terms
of a,
b}
c.
Reasons
Argument
the altitude
ha denote
1. Let
jectionof
A
b
upon
a, and
upon
T
a, p
the
the proarea
1. " 485.
of
ABC.
Then
T=-aha
2.
ha2=b2-p2
3.
Butp
-ha.
=
'
2
2
a
2. " 447.
(b+p)(b-p).
=
a2 + 52-c2
=
2
*-"?
-"* +
or
(Fig.
1),
"
K
"
2
a
3. " 456.
a
(Fig.2).
4..-.K
a2 + b2
rX-*^
"4
2a6
4. " 309.
+ c2
2a"-a2-"2
+ a2-h"2-c2
2a
2a
c)(c+a" 6)(c"a+6)
6"
(a+6+c)(q-f
=
4a2
/(a+"+ c)(q+ft-c)(c+a-6)(c-a+6)
,
k
o.
K=y"
_^
.
.
6. Now
Then
let
_
a
+ 6 +
a
+ "-c
a-"
6 +
+
c-a
c
c
=
2
=
2s-
=
2s-2"
=
2s-2a
s.
2c
=
=
=
2(s
"
c);
6);
2(s
2(s-a).
"
t.
5. "
54, 13.
6. "
54, 3.
221
IV
BOOK
Reasons
Argument
2
K=yj 2(s-a)
'.Then
=
Vs(s
-
/.
T=-
"
V"(s
Find
Ex.
834.
If the
the
In
835.
of
area
sides of
altitude upon
c.
triangleABC,
a
The
8, b
=
7, 10, and
are
a,
12, c
=
=
16 ; find
b ; the altitude upon
product of its
the
formula
of
area
c.
of a triangle is equal to one
perimeter and the radius of the
area
the
13.
Theorem
V.
Proposition
491.
sides
6, and c, write
(See Prop. IV, Arg. 7.)
triangleare
a
b ; upon
trianglewhose
a
; the altitude upon
triangleABC
Q.E.F.
"
"
833.
Ex.
8. " 309.
-
-
ft)(s ")(s c).
"
Ex.
for the
"
ft)(s b)(s c)
-
=
4 ft2
"
Vs(s
-
" 309.
7.
d)(s b)(s c).
"
ft
8.
2(8-6) 2(s-c)
s
the
half
inscribed
circle.
b
A
A
Given
circle
inscribed
To
with
ABC,
1
of A
about
of A
OBC
OAB"^c-r.
492.
Cor.
a
pU'"l by
radius
of
b -f-c)r.
+
(ft
Outline
1. Area
c, and
r.
T=
prove
T, sides a, b, and
area
The
\
=
2.
area
a
"
.'. r
of
of the
area
r\
=
circle is -equal to
the radius
Proof
of
of A
OCA
=
i
6
.
i(ft-r-"-fc)r,
any
one
polygon
half
inscribed
its
area
q.e.d.
circumscribed
perimeter
circle.
r;
multi-
222
PLANE
If the
836.
Ex.
3, 5, and
are
inches,find
7
Derive
837.
Ex.
of
area
GEOMETRY
triangleis
a
formula
a
for
of the
of the sides
trianglein terms
Derive
838.
Ex.
about
T="(a
triangle,in
a
Outline
1.
dh
of
and
its sides
circle inscribed
a
$ (2 s)r
=
in
a
radius
sr.
=
of
triangle.
2 i2
or
circle circumscribed
a
(See figure for
i.e. d
ac;
=
circle.
radius
the
Solution
of
inscribed
sides of the
of the
terms
c)r
for
formula
a
inches
Solution
of
b +
+
the
square
triangle.
Outline
1.
of the
radius
the
V3
15
=
798.)
Ex.
"
.
h
2.
,'.M=
Q.E.F.
=
2 h
If the
839.
Ex.
radius
of
kind
radius
of the
tb2=ac
4.
.-.
-5*-.
=
+
a
ac(a
b +
+
(a
8.
rs.
"
Likewise
2.
But
5.
Likewise
the
r,
a
Ex.842,
the radius
Find
a.
circumscribed
the
circle.
by comparing
the
longest side.
bisectors
:c=r:s.
8
-^-.
=
a
cy
(a
+
y/bcs(s
+
of the
angles of
a
triangle
797.)
.-.
a
6.
.-.
tf
-\-c
: a
=b
: r.
ahH
ac-
=
(a-\-c)2
c
=_^_Vacs(s
%
...
_
K
"
a
tc =
a
and
Ex.
3.
7"
cy
a) and
"
B, T, ha, ma,
the
5
and
answer
your
b+c
Find
of
(See figure for
c
ta =
11, find
circle.
4 a,
3a,
c)(a-b+c)=4acs{s-bK
+
and
c)
-
triangle.
Solution
of
b) 0
-
circumscribed
Verify
for the
formulas
Derive
Outline
1.
it ?
(s
9, 10,
radius
circle with
of the sides of the
in terms
r
circle ; the
circumscribed
841.
of the
triangle are
a
triangleis
a
V s(8 -a)
triangleare
a
radius
of
inscribed
the
What
Ex.
sides
The
of
of
circle ; the
of the inscribed
Ex.840.
sides
4
+
6
+
Vabs(s
K
"
c)
J
"
Q-E-F-
ta, having given :
a
=
11,
b
=
9, c=
16.
Ex.
848.
a
=
13,
b
=
15,
c
=
Ex.
8,44.
a
=
24, b
=
10, c
=
What
kind
of
an
6).
J
c
angle is
C?
20.
What
kind
of
an
angle is
C?
26.
What
kind
of
an
angle is
C ?
BOOK
TRANSFORMATION
493.
To
Def.
figurewhich
is
OF
transform
equivalentto
To
construct
FIGURES
figure means
a
find
to
VI.
Problem
triangle equivalent
a
another
it.
Proposition
494.
223
IV
to
a
given
polygon.
polygon ABGDEF.
Given
To
construct
Construct
(a)
polygon ABCDEF.
polygon =c= ABCDEF,
A
a
a
=o=
I.
1.
Join
any
two
2.
Construct
BG
3.
Draw
4.
Polygon
alternate
vertices,as
C
and
A.
prolonged at
FA
GCDEF
=o
and
polygon ABCDEF
has
and
AGC
CA, and
the
A
But
.-.
AGC
=c=
A
polygon A
the
base
same
the
altitude,
same
lisCA
and
polygon GCDEF
"
o=
side less.
1.
"235.
2.
"486.
3.
By iden.
"54,2.
J_ between
BG.
ABC.
CDEF
one
Reasons
have
ABC
the
.-.
" 188.
Proof
Argument
A
G.
CG.
II.
1.
side less.
having one
Construction
meeting
II CA,
but
polygon
ACDEF.
polygonAB
CDEF.
Q.E.P.
4.
224
PLANE
In like
(6)
polygon
the
reduce
manner,
until A
GCDEF
GEOMETRY
new
discussion
left
are
as
exercise
an
student.
for the
Ex.
845.
Transform
a
scalene
Ex.
846.
Transform
a
trapezoid into
Ex.
847.
Transform
a
parallelogram into
Ex.
848.
Transform
a
pentagon into
Ex.
849.
Construct
Ex.
850.
Transform
Ex.
851.
Construct
and
of sides of the
is obtained.
DHK
construction,
proof,and
The
number
which
have
a
a
The
and
rhomboid
and
altitude
one
a
triangle.
trapezoid.
a
isosceles
of
into
rhombus
triangle.
a
a
given trapezium.
triangle.
which
are
equivalent,
diagonal.
common
area
an
isosceles
righttriangle.
given pentagon
a
VII.
Proposition
495.
a
an
triangleequivalent to "
$ of
a
triangleinto
Theorem
of a trapezoid equals the product of its
half the sum
of its bases.
V
Given
To
trapezoidAFED,
prove
area
D
b
A
of
AFED
with
=
altitude
I- (h 4-
h and
b^h.
bases
b and
b1.
226
GEOMETRY
PLANE
Proposition
VIII.
Theorem
Two
have an
triangles which
angle of one equal
to an
angle of the other are to each other as the products
of tTie sides including the equal angles.
498.
G
A
A
Given
To
and
ABC
with
BEF,
A
ABC
AC
"
A
BEF
BF
"
Z
A
=
Z.B.
AB
prove
BE
Reasons
Argument
1. Let
the
be
h
altitude
side
*
*
2.
In
3.
.'.
4.
AC-h
A
BF
A
ABG
h
_AB^
~
A
"
h'
ZA
BEK,
AC
AB
A
BEF
BF
BE~
one
shall
of the
sides
the numerical
2.
By hyp.
3.
" 422.
4.
"
BEF
=
ZB.
BEK.
ABC
859.
" 487.
hr
BF
A
Ex.
1.
ACmh_
and
ABG
of A
upon
h'~~ BE'
5.
of
A ABC
rt. A
ABC
Then,
BF.
BEF
A
h' the altitude
side AC, and
upon
of
_AC
"
AB
Q.E.D.
BF
424,2.
5. " 309.
-BE
trianglesupon the blackboard, so that
Give a rough estimate
equal an angle of the other.
and
includingthe equal angles in the two triangles,
Draw
ratio
two
of the
an
angle
in inches
compute
triangles.
supplementary to
triangleshave an angle of one
other
the products of
as
trianglesare to each
angle of the other,
sides includingthe supplementary angles.
Ex.
860.
If two
the
an
the
Proposition
To
499.
A
Given
To
construct
with
ABC,
construct
a
a
IX.
square
b and
A
=0=
Problem
equivalent
square
base
227
IV
BOOK
to
given
a
angle
tri-
altitude h.
ABC.
I. Analysis
1.
of
Let
#
the
=
side
of the
required
"c2=
then
square;
area
required square.
2.
\
h
"
.\x2
3.
4.
5.
b
\b
.*.
: x
the
of the
=
area
=
"b"h.
=
x
side
:
i
it x.
Construct
a
b and
mean
will
required square
II.
1.
ABC.
h.
of the
between
given A
be
a
Jb
and
mean
h.
Construction
proportionalbetween
h.
" 445.
2.
On
x,
3.
"S
is the
as
portional
pro-
base,construct
required square.
a
square,
S.
Reasons
Q.E.D.
1.
By
2.
" 388.
3.
" 478.
4.
" 485.
5.
" 54, 1.
cons.
Call
228
GEOMETRY
PLANE
discussion is left
The
IV.
500.
b and
501.
Could
Question.
|h
x
as
constructed
he
for the student.
exercise
an
as
a
tween
proportionalbe-
mean
?
To
Problem.
construct
a
equivalent
square
to
a
given parallelogram.
Ex.
861.
Construct
a
square
equivalentto
a
given rectangle.
Ex.
862.
Construct
a
square
equivalentto
a
given trapezoid.
Upon a given
given parallelogram.
to
Ex.
863.
Ex.
864.
Construct
a
base
construct
rectangle having
a
given base
a
the
Props. VI, VIII, and IX form
class of important constructions.
(a) Prop. VI enables us to construct
any
polygon.
trapezoid,an
or
and
to
a
equivalent
given square.
a
502.
to
triangle equivalent
a
rhombus
It is then
an
easy
a
basis
of
large
a
triangleequivalent
matter
to
construct
a
isosceles
trapezoid,a parallelogram,a rectangle,
equivalentto the triangleand hence equivalent
to the
given polygon.
for constructingan
lateral
equi(b) Prop. VIII gives us a method
triangleequivalentto any giventriangle.(See Ex. 865.)
Hence
an
Prop. VIII, with Prop. VI, enables us to construct
triangleequivalentto any given polygon.
equilateral
struct
(c)Likewise Prop. IX, with Prop. VI, enables us to cona
equivalent to any given polygon or to any
square
fractional part or to any multipleof any given polygon.
into triangleDBC,
retaining
triangleABC
(a) Transform
60".
DBC
base BC
and making angle
into
Transform
triangle EBF,
retaining angle
triangle DBC
(6)
and BF
60" and making sides EB
DBC
equal. (Each will be a mean
and BC.)
DB
proportionalbetween
kind
of a triangleis EBF?
(c) What
865.
Ex.
"
=
Ex.
Ex.
866.
867.
trapezium.
Transform
Construct
a
an
parallelogram into
an
equilateraltriangle.
equilateraltriangleequivalent to f
of
a
given
BOOK
Construct
229
IV
of the
following figuresequivalentto f of a
given irregularpentagon : (1) a triangle; (2) an isosceles triangle; (3) a
righttriangle; (4) an equilateraltriangle; (5) a trapezium ; (6) a trapezoid
isosceles trapezoid ; (8) a parallelogram; (9) a rhombus
; (7) an
;
868.
Ex.
(10)
a
each
rectanglej (11)
Ex.
having
a
a
given line
as
trapezoid into
given line.
Construct
870.
a
square.
Transform
869.
hypotenuse equal to
given line ?
Ex.
a
and
a
Proposition
503.
squares
Two
similar
of any
two
right trianglehaving the
restrictions
there
are
upon
the
given trapezoid,and
given angle adjacent to the base.
triangleequivalent to
a
base
What
a
X.
a
Theorem
to each
triangles are
homologous sides,
other
as
the
B
Given
homol.
similar
two
A
ABC
and
A'b'c',with,
b and
b' two
sides.
Reasons
Q.E.D.
A
A'B'c'
b'
b'
b'2
1.
By hyp.
2.
" 424, 1.
3.
" 498.
4.
"
5.
" 309.
424, 2.
230
PLANE
504.
of any
squares
Ex.
Ex.
four
Ex.
an
a
875.
Prove
Ex.
876.
Draw
fourths
trianglesimilar
a
X
Prop.
In
line
two
a
given triangleand
that of
a
given
equivalent parts by
two
the
base
one
third of the
parallelto
an
line
a
of
triangleand cuttingoff
given triangle.
a
equilateraltrianglewhose
b',etc.,are
a
are
similar
10
angle
tri-
two
-=-2.
shall be three
polygons
homol.
are
to
each
other
as
the
and
a\b
sides.
homologous
pairsof
a2
area
Theorem
XI.
polygons
similar
two
sides
square.
similar
of any
prove
having
their difference.
P
To
to
similar
Construct
Given
and
given triangleand having
487.
equivalent to
878.
Two
squares
of
squares
a
by using "
Proposition
505.
the
as
to
given triangleinto
a
a
equivalent to
Ex.
other
trianglesa pair of homologous
feet,respectively. Find the homologous side of
877.
6
trianglesimilar
a
trianglethat shall be
feet and
to each
(See " 435.)
base.
Ex.
Ex.
trianglesare
the
as
great.
as
Divide
the
parallelto
altitudes.
other
great.
as
874.
Ex.
to each
medians.
Construct
twice
area
similar
times
873.
triangles are
homologous
Construct
872.
area
similar
two
Two
871.
homologous
two
an
Two
Cor.
GEOMETRY
P
and
sides.
P' in which
a
231
IV
BOOK
Reasons
Argument
homol.
two
the
II and
I',A
each
A
placed,as
similarly
" 439.
into
be divided
~
" 54, 15.
any
V1.
and
V
of A
number
same
each and
A
as
vertices,
polygonswill
the
Then
2.
possiblediagonals from
all
1. Draw
to
I and
II',etc.
A
AI
Then
3.
I'
A
All
4.
Air
A
III
A
III'
5.
a_
But
6.
'd'2'
1
"
"
a'' e'~dr
t.=JL
7.
**
a'2' e'2
II
A
8.
"
9.
AT'
d'2'
Air
A
I + A
11+
A
I' + A
II' + A
III
A
III'
a2
_AI
III
A
A
.
'
*
10.
of any
squares
Ex.
Q.E.D.
a'2'
similar
Two
Cor.
,'2'
I'
=
'
P'
506.
A
?L
L
"
'
III'
polygons are to each
homologous diagonals.
two
879.
If
880.
Divide
one
square
is double
another,
what
other
as
the
is the ratio of their
sides ?
Ex.
one
part shall be
Ex.
881.
the squares
Ex.
ogous
a
The
of their
882.
s^de of
One
a
a
given hexagon
hexagon
of
areas
similar
two
into
to the
similar
equivalentparts
two
so
that
given hexagon.
rhombuses
are
to
each
other
homologous diagonals.
side of
similar
a
polygon
polygon
is 8 and
its
area
is 12 ; find its area.
is 120.
The
homol
as
232
GEOMETRY
PLANE
Proposition
The
507.
XII.
described
square
on
the other
Given
rt. A
To
to
the
sum
sideSo
two
at C, and
right-angled
ABC,
AD
square
prove
the squares
=c=
BF
square
+
square
Argument
1. From
draw
C
and
and
CK
BCA, and
4.
and
5. In A
GCB
and
CAK
FCB
are
HAB,
6.
Z
7.
Abak=Zhac.
'
8.
9.
.*
.
Z
CAB
CAK
=
=
.*. ACAK=
10. A
CAK
base
"
cuttingAB
at L
BH.
3. A ACG,
.'. ACF
" AB,
CM
at M.
KD
2. Draw
hypotenuse of a
of the squares
sides.
its three
on
the
on
right triangle is equivalent
described
Theorem
str. lines.
CA
Z
CAB.
Z
HAB.
A
HAB.
rectangleAM
AK
and
_L between
the
HA,
=
and
the
all rt. A.
are
have
the
=
AB.
same
altitude,the
same
lisAK
AK
and
CM.
CH.
described
234
PLANE
510.
Historical
Prop.
Note.
it
Proposition,because
is that
here
given
GEOMETRY
of Euclid
is usuallyknown
XII
discovered
was
(about 300
the
as
by Pythagoras.
rean
Pythago-
The
proof
they
knew
B.C.).
Pythagoras (569-500 b.c),
famous
of the most
one
of
antiquity, was
of manhood
early years
under
in Asia
Thales
India.
where
he
that
returned
he
Southern
all
their
reverent
many
They
five-pointedstar
common.
badge
and
the
and
of which
have
suffered
who
attributed
kept
were
shipwreck
on
his death
and
a
a
prove
Ex.
to
the
884.
the
the
Pythagorean
885.
sum
Use
Construct
of two
Deity.
of
the hatred
murdered
a
by
after
hundred
adjoining figure to
theorem.
triangleequivalent
B
given triangles.
a
important
and
numbers
since
sacrilege,
of the
years.
Ex.
most
theory of incommensurables
of the
over
their
coveries,
dis-
proved
incommensurable.
are
square
reestablished
was
it flourished for
him
secret.
account
thought to be symbolical
Pythagoras, having incurred
his politicalopponents,
was
his school
of
to
incommensurable
of
propounded
were
them, but
which
They
they considered symbolical of health.
severe
practiced
discipline,
having obedience,temperance,
their ideals.
brotherhood
The
regarded their leader
esteem
first man
pentagram
or
which
Pythagoras knew something
that the diagonal and the side
The
Pythagoras
possessed
simple food and
and purity as
with
He
society,the
to construct
ate
there
closest followers,
of which
as
how
his
secret
things in
used
and
in
adherents.
gained many
formed, with
members
success.
Crotona
to
Italy
a
school
a
great
a
went
and
to Samos
established
not
was
Later
and
Egypt
Babylon
also in
He
ing
studytraveled
and
and
Minor
probably
spent his
He
at Samos.
born
maticians
mathe-
such
is saidto
numbers
The
886.
Ex.
Make
accurate
figurerepresents
the
of
in
acres
Ex.
dimensions
=
90
and
a
rods, WX
XZ
=
1 inch
scale
calculate
farm
as
40
100
rods,
Draw
the
to
rods,and
of
area
I
the
1
I
TC0
40
20
the rectangle
By
Def.
of
is
lines
two
rectangle having these
the
lines
two
adjacentsides.
The
888.
Ex.
two
ZW
rods,
acres.
511.
meant
XF=60
farm
=
a
ing
follow-
the
=
the
in
of
66 rods.
of the
plot
farm
form
:
70
=
the scale indicated.
proximately
ap-
A
trapezium, has
YZ
to
farm.,
the
in the
rods,
drawn
farm
number
887.
XYZW,
a
ments
measure-
calculate
and
235
IV
BOOK
lines is
described
on
difference
The
of
the
of
sum
equivalent to the sum
the lines plus twice
889.
Ex.
diminished
lines
twice
AB
their
squares
rectangle.
the
on
the
to
the
on
of
sum
lines
rectangle.
their
and
the
of
equivalent
described
squares
by
is
the
on
described
square
two
Let
Hint.
described
square
be
CB
the
given lines.
li
Ex.
890.
and
sum
difference
equivalentto
on
the
the
whose
respectivelyof
difference
sides
two
of the squares
are
the
lines
is
described
lines.
Hint.
Ex.
rectangle
The
Let
891.
last three
AB
Write
exercises.
and
BC
the
be the
three
given lines.
algebraic formulas
B
corresponding
to
C
the
GEOMETRY
PLANE
236
512.
To
construct
given
two
the
to
of
sum
squares.
Q
p
Given
To
equivalent
square
a
Problem
XIII.
Proposition
P
squares
construct
a
and
Q.
square
the
=c=
sum
of
and
P
Q.
I. Construction
1. Construct
the
rt. A
ABC,
having
for its sides p
and
g,
the sides of the
given squares.
On r, the hypotenuse of
R is the required square.
2.
3.
II.
proof and discussion
The
Construct
892.
Ex.
the
a
A,
R.
left to the student.
are
equivalent to the
square
the square
construct
of three
sum
or
more
given squares.
and
Ex.
893
Ex.
894.
a
equivalentto
square
the difference
equivalentto the
a
square
a
polygon
sum
of
of two
a
squares.
given
square
895.
Construct
896.
their
Construct
similar to two
(See " 509.)
sum.
polygon similar
a
given similar polygons
to two
given
similar
polygons
equivalentto their difference.
Ex.
two
Construct
equivalentto
Ex.
and
Construct
given triangle.
a
Ex.
and
.
897.
Construct
an
equilateraltriangleequivalent to the
sum
of
given equilateraltriangles.
Ex.
898.
of two
Construct
an
ence
equilateraltriangleequivalent to the differ-
given equilateraltriangles.
BOOK
XIV.
Proposition
To
513.
To
construct
and
Q, with
and
polygon
a
similar
a
polygons P
Given
Problem
polygon
equivalent
construct
given polygons
237
IV
and
P
~
side of
a
s
of
one
two
other.
the
to
to
P.
Q.
=c=
I. Analysis
the
Imagine
1.
polygon with
2.
Then
3.
Now
we
P
reduce
squares
be
m
4.
.*.
m2
:
5.
.\
m
6.
That
514.
Ex.
to
P
n2
=
is,x
s
and
Q to
:
s2
=
:
x2,since
similar,
not
are
(1)
R.
o
the sides of these
Let
m2
Q
=o=
P
and
n2
=o=
Q.
(1).
: x.
is the fourth
to
proportional
construction,
proof,and
Note.
m,
discussion
n, and
s.
left
are
as
an
This
problem
first solved
was
by Pythagoras
b.c.
Construct
trianglesimilar
given parallelogram.
to
a
Construct
a
Ex.
901.
Construct
a
Ex.
Q
squares.
o=
x2,from
900.
a
:
the required
side of P.
; then
71, respectively
Ex.
that of
x2 ; i.e. P
a
be
R
for the student.
899.
a
:
to s,
let
and
comparing polygons which
s2
=
Historical
550
s2
=
and
: n
The
exercise
about
R
homol:
x
to avoid
may
II.
side
:
solved
problem
square
a
given triangleand equivalent
equivalentto
a
given pentagon.
triangle,
given its angles and
See Prop. XIV.
given parallelogram). Hint.
902.
perpendicular
Prop.'"IV.
Divide
to the
a
triangleinto
base.
Hint.
two
Draw
its
area
equivalentparts by
a
median
to
a
(equal to
line drawn
the base, then
apply
238
GEOMETRY
PLANE
Ex.
1
Fig.
903.
scale
the
represents
indicated.
of
maps
Colorado
and
Utah
drawn
to
By
the
carefully measuring
(1) Calculate the
maps:
state.
perimeter of each
(2)
Calculate
each
state.
your
results
the
Check
(3)
for
with
comparing
given for
U
COLORADO
T
(2) by
the
these
of
area
areas
in
states
geography.
your
SCALE
Ex.
904.
sents
a
nia.
A
the
Fig.2 repreof
map
300 miles
OF
MILES
300
400
FlG
j
from
the northeast
to
corner
is
corner
long.
^s\
Determine
(1)
200
Pennsylva-
straightline
southwest
"?
50
scale to
the
which
the
map
Scranton
"
is drawn.
(2) Calculate the distance from Pittsburgto
Harrisburg ; from Harrisburg to Philadelphia;
from
Philadelphiato Scran ton ; from Scranton
Harrisburg.
(3) Calculate
to
1
a
square
and
screen
3
from
0 represent
the
B
a
Harrisburg
area
*
Philadelphia i
2.
of the
A
and
feet
C ; D
3
and
square
screen
dle
can-
2
screen
from
yards
feet
a
foot square
1
C;
2
If
C.
approximately
screen
from
yard
Pittsburg
Fig.
Let
905.
; A
.
miles.
state in square
Ex.
PENNSYLVANIA
a
yards
removed,
were
quantityof lightit received would
fall on B.
What
would
happen if B
the
were
removed
then,
would
intense
the
?
law
Ex.
from
AB
the
From
Let
AC
=
AM
"
AX,
two
3.
equivalent parts by
a
line drawn
of its sides. "
one
be
M
Fig.
triangleinto
a
given point in
"
screen,
light be the least
determine
the figure,
Divide
906.
a
which
of intensity of light.
Hint.
i
On
?
the
given point
where
MX
in
is the
AC
of
triangle ABC;
required line.
then
BOOK
Ex.
A
907.
track BA
and
represents
with
given radius
intersectingcar lines,
DE,
line
is tangent to
This
a
the
buildingbetween
point
its
that
so
Find
the
two
each
car
designing
forming at their
small
acute
angle, as
New
York
City.
a
the
of
area
rhombus
a
that
Show
909.
polygon
if
sv
if
their
EXERCISES
and
a
sum
"
sides of
two
are
a
triangle,the
of the
areas
in
chords
a
arm
Ex.
913.
of the
within
a
consider
triangles.
on
segments of
the
on
square
two
perpendicular
the diameter.
hypotenuse of a righttriangleis 20, and the projection
is its area?
hypotenuse is 4. What
quadrilateral
the
angle included
and
the
included
is
between
angle of the
Transform
equivalent to a triangleif its diagonals
them
are
respectivelyequal to two sides
triangle.
given triangleinto another
a
trianglecontaining
given angles.
two
be +
+
Prove geometricallythe algebraicformula
915.
Ex.
ac
point
the
A
914.
any
120".
polygon, and
of the
vertices
equivalentto the
and
Ex.
the
of the squares
sum
The
upon
one
point with
circle is
912.
Ex.
of
The
perpendiculars from
included
the
or
is
area
sides is constant.
the
upon
the
911.
Ex.
of the
sum
Join
Hint.
the
The
910.
convex
circle
a
A"'*
the included
angle is 30" or 150"; \ abVll when
" ab when
angle is 45" or 135"; \ ab V3 when the included angle is 60"
Ex.
on
FlG- 4MISCELLANEOUS
Ex.
of
arc
an
is 16.
sum
station
in
in the ratio of 5 to 7 and
diagonalsare
Construct
streets
two
Building in
908.
approach the
arc.
intersection
the Flatiron
Ex.
and
principleis involved
same
of
track AC.
on
connecting
r,
Cars
station.
a
leave the station
239
IV
Ex.
ad +
the
angle ("69), then
of the squares
Ex.
at
917.
P.
=
bd.
trianglean
If in any
916.
(a + 6) (c -f d)
on
the
The
two
medians
two
that the
Prove
on
square
other
angle is equal to
two
thirds of
a
straight
oppositeis equivalentto the sum
sides and the rectanglecontained
by them.
the side
BK
and
triangleBPS
sect
intertriangleB8T
equivalentto the quadrilateral
SH
is
of the
HPKT.
Ex.
918.
Find
andv7inches and
the
the
area
included
of
a
triangleif
angleis 30",
two
of its sides are
6 inches
210
PLANE
Ex.
919.
By
trianglewhose
Ex.
GEOMETRY
different
two
methods
find the
of
area
equilateral
an
side is 10 inches.
920.
The
of
area
equilateraltriangleis 36 VS
an
; find
side and
a
altitude.
an
Ex.
921.
formula
Ex.
C is
a
Ex.
whose
for the
922.
of
area
What
formula
of
Prop. IV, Arg. 8, derive
equilateral
trianglewhose side is a.
the
'
an
does
the formula
for T
in
Prop.
IV
become
if
angle
rightangle ?
923.
Given
is O.
center
triangleODC
924.
the
G
vertex
Draw
D.
at
is
equilateraltriangleABG,
an
At
the circumference
Ex.
the
By using
the
erect
inscribed
in
perpendicularto
a
radii OD
and
OG.
circle
a
BG
cutting
Prove
that
the
triangleswhich
have
an
equilateral.
Assuming
that
the
of
areas
two
angle of the one
equal to an angle of the other are to each other as the
products of the sides including the equal angles,prove that the bisector
of an angle of a triangledivides the opposite side into segments proportional
to the adjacent sides.
Ex.
925.
rhombus
A
altitude of the rhombus
and
a
have
square
is three fourths
equal perimeters, and
its side ; compare
the
the
of the
areas
figures.
two
Ex.
926.
from
the radius
Ex.
length
The
the
subtending
927.
In
any
928.
and
the
hypotenuse
thus
929.
The
each
of
bases
of the
how
A
930.
a
of
area
the greatest perpendicular
is 2 feet
is 5 inches, find the
If the
ratio
931.
Construct
Ex.
932.
If the
933.
Divide
a
inches.
7"
a
base
area
one
of the
the vertex
Find
given triangleinto
the given triangle.
feet and
16
is 5 feet.
trapezoid,if
of 8 inches
the
Find
of its
a
base
largertriangle.
similar
Hint.
greater?
equivalentto
one
third of
a
areas
two
to
1,
"" 418, 503.
given square.
equilateraltriangleis equal
ratio of their
parallel
triangle
is 7
triangles
See
non-
line
smaller
of the
of two
of the
area
each
is cut by
of the
angles
tri-
two
feet,respectively,
10
If the
righttriangleinto
rightangle
the
similar
in the
square
is the
a
a
of similitude
side of
another, what
from
divides
it.
from
6 inches
Ex.
Ex.
chord
non-parallelsides
often is the less contained
of
feet,and
trapezoid are
trianglehaving
base and
formed
Ex.
the
similar to
and
other
trapezoid. Also find the
parallelsides is 8 feet.
to the
to
arc
is 10
a line
righttriangle
similar to each
Ex.
chord
a
of the circle.
perpendicular to
Ex.
of
to
the
?
isosceles triangles.
tude
alti-
242
PLANE
feet of the
The
942.
Ex.
triangle from
point
any
Outline
AP
as
circumference
diameter
will pass
andZ2
=
of
the
upon
the sides of
circumscribed
a
circle
.:
Z PKM
=
Z PBM.
ZAPB
=
ZQPK.
=
ZBPK.
=
Z BMK.
.-.
other
and
"M
Ex.
AMQ
Z
.-.
Given
943.
sides of
two
Analysis.
ABC
-f
a
form
MK
the
a
Imagine
Since
given.
str.
base,
the
Q.
line.
the
angle at
triangle;construct
Prolong AB,
is
one
and
Similarly
Z2'.
Z3',
..ZAPQ
having
M
through
=
Z3
and
circle
The
Proof.
of
Zl=ZV
.-.
c
perpendicularsdropped
in the
collinear.
are
A
GEOMETRY
problem
making BE
ZE=\ZABQ,
=
the
the
vertex, and
A
sum
of the
triangle.
solved
BC,
the
and
since
draw
the
AEG
line
be
can
constructed.
Ex.
in
944.
magnitude
of the
Ex.
one
and
position;
947.
a
righttriangleis given
find the
locus
of the center
circle.
Prop. VIII, Book IV, by using the following figure,
is placed in the positionABE.
A'D'E'
946.
of
Prove
945.
will not
Ex.
hypotenuse
inscribed
in which
Ex.
The
Prop. VIII, Book IV, using
fall wholly within the other.
Prove
If two
triangleshave
other, the trianglesare
an
angle of
trianglessuch
two
one
supplementary to
the products of
other
to each
as
angle of the
sides including the supplementary angles. (Prove by
that of Ex, 945.)
that
method
an
the
similar to
angles ;
Add
(b
Z 1 to
Zl
+
.-.
ZB
The
Z2
"
both
2Z1
A
show
that
ZA.
+
base
be
now
may
Zl
.-.
-)-ZA.
Zl
=
then
equation ;
ButZl
ZA.
+
Z2
of the
members
2 Zl
=
=
A BE
of
accompanying figuresuppose
"), to be given. A consideration
figure will
the
or
difference
sides, and
In the
EA,
and
of
triangle.
the
construct
Analysis.
c
base, difference
Given
948.
Ex.
243
IV
BOOK
Z2
+
l(ZB
=
constructed.
ZB.
=
ZA).
-
The
is left
of the construction
rest
student.
for the
Given
949.
Ex.
base, vertex
angle, and difference of sides,construct
AB-BC.
ZAEC
triangle.
the
AE
Analysis.
.
A
=
AEC
Ex.
If upon
950.
constructed.
be
can
.
sides
the
of any
constructed,
equilateraltrianglesare
90"
triangle
lines
the
of the equilateraltriangles
joining the centers
form an equilateral
triangle.
Hint.
circles about
Circumscribe
equilateralA.
Join
intersection
of the three
Prove
Z
and
each
Ex.
supplement
951.
Analysis.
the
inscribe
To
Imagine
a
C,
Z
opposite
in
a
given A.
equilateralA,
the vertices of the
in the
in
A
to
an
lateral.
proved equi-
be
semicircle.
_AB
OA
a
opposite
problem solved
the
Prove
required square.
the Z
square
and
of the
supplement
of
three
point of
common
to A, B,
circles,
0 the
at
also the
the
0,
the
and
Draw
.
ABCD
\B
EF"
EO
BA:
AO.
2
meeting OB
Ex.
FE
F.
at
:EO
=
E
OE.
EF=2
.-.
prolonged
To
952.
inscribe
a
square
in
A
0
D
given
a
triangle.
Outline
solved
Draw
SFW
Draw
BF,
BC
and
^
'gD.
Imagine the problem
the
ABCD
and
BT
will be
cons,
Solution.
of
required
construct
square.
KSFH.
square
thus
determining point C.
evident from the figure. To
BO
:
SF
=
CD
:
FIT.
But
SF
KD
T
H
The
prove
=
FH.
ABCD
a
.:
BC
square,
=
CD.
prove
244
so
In
Ex.
953.
that
its vertices
ABC
is
would
formed
be
Then
A
required
Ex.
its two
and
AKT
KT
be
can
a
given side,
given square.
mc,
solved
and
sum
of
arms.
/i".
and
moved
were
the
K,
having
square,
triangle,given ma,
a
itself till it contained
to
a
A, given the hypotenuse
rt.
a
Imagine the problem
If BF
the required A.
Analysis.
that
construct
square
shall lie in the sides of the
Construct
954.
Ex.
given
a
Construct
Hint.
II
GEOMETRY
PLANE
AKT
A
rt.
would
equal \ BF.
made
the
of the
basis
construction.
Construct
955.
diagonals,and
Outline
the
Construction.
op
solved
problem
quadrilateral,
given
between
them.
angle
a
that
and
Imagine
the
is
the
ABCD
of its opposite
two
sides,
required quadrilateral,
s, s',d, d',and Z DOC
of d and
d' the
being given. By II motion
be obtained.
The
parallelogram BKFD
may
ing
required construction may be begun by drawO
as
are
With
an
intersectingthe
locate
Ex.
sides
two
known.
radius,describe
arc
to
since
BKFD,
Z
K
and
first,
D
s
and
center
as
C.
at
as
cluded
in-
and
center
as
; with
arc
the
Construct
EJ
sr
as
radius,describe
ABKC,
or
ED
an
DACF,
A.
Between
956.
the line of centers
two
and
circles draw
equal to
a
line which
a
given line
shall be
parallelto
Z.
/
Find
Ex.957.
=^,
x
where
a,
b, c, and
d
represent given lines.
d2
Here
Hint.
by
a
method
Then
d
^VS
=
=
the
and
construct
given triangleinto
any
that used
.-.
in Ex.
of the
side of the
~b-h.
2
ab
y.
Then
construct
x.
d
Call the base
x
4
d
Let
different from
Analysis.
Let
C
Transform
958.
Ex.
ab
x
equilateraltriangle
8G5.
given triangleb and its altitude
required equilateral
triangle.
lb:x
*
an
=
x:hV3.
a.
V
BOOK
OF
MEASUREMENT
POLYGONS.
REGULAR
THE
CIRCLE
515.
and
is
A regular polygon
Def
.
is both
which
one
equilateral
equiangular.
Ex.
959.
Draw
an
Ex.
960.
Draw
a
equiangular but
a
regular polygon ?
that is equilateral
but
quadrilateral
equiangular ;
not
equilateral; neither
not
equilateralnor equiangular ; both
of these
quadrilateralsis a regular
equiangular. Which
equilateraland
Is it
equilateraltriangle.
polygon?
Find
961.
Ex.
516.
2"
the
3,
"
2n
5, 2n
"
nineteen
n
by
(2n + 1)
condition
all
Gauss
can
be
that
and
known
In
a
*
it
Gauss
product of
two
be
divided
of 17 sides
that
2n,
by Gauss,
be
can
structed
con-
integer
an
satisfying this
polygons having
different
more
into
general it is possible
numbers
or
The
arcs.
later.
sides,n being
proved also
the possibility
angular magnitude
that in
first four
decagon.
regular do-
elementary geometry,
discovered
was
and
compasses,
a
numbers
constructed.
that
only
a
by elementary geometry.
" 52.0.
*
the
that
1796
equal
discussed
regular polygon
The
number.
proved, moreover,
constructive
will be
polygons having (2" +1)
prime
of
methods
the
which
was
of
number
a
a
presupposes
circumference) could
a
3, 5, 17, 257.
sides equal to the
of this series
see
of age,
are
of
number
a
it
time
of ruler
means
by
equal angles.
15
years
construct
to
and
"
obtained
hence
point (and
a
into
specialcases
Euclid's
early as
angle of
an
following theorem
The
be
cannot
in certain
except
about
degrees in
dividing the circumference
division
As
of
number
Note.
Historical
of
actual
the
245
class of
limited
For
a
note
on
regularpolygons
the
life of
are
Gauss,
246
PLANE
GEOMETRY
Proposition
I.
Theorem
If the circumference of a circle is divided
number
of equal arcs : (a) the chords joining
any
points of division form a regular polygon inscribed
517.
the
circle;
form
a
(b) tangents
regular polygon
circumference
Given
let chords
etc.,and
and
division,
To
prove
AB,
let the
at the
ACE
"
at
the
circumscribed
in
points of
division
the
circle.
about
equal arcs AB, BC, CD.
etc.,join the several points of
into
divided
BC, CD,
the
ference
tangents GH, HK, KL, etc.,touch the circum-
several
A BCD
drawn
into
"
points of
"
and
GHKL
division.
"
"
"
regularpolygons.
247
V
BOOK
Reasons
Argument
11.
.\
GH=^HK=KL=-
12.
.-.
GHKL
518.
is
"""
?
contain
is divided
into
n
"54, la.
12.
" 515.
figure of " 517, the circumference
equal to arc AB,
many
arcs, each
how
How
?
DFB
arc
11.
the
If,in
equal parts,
six
-.
regularpolygon,q.e.d.
a
Questions.
into
CEA
"
will
many
each
In step
equal parts ?
of
contain
will
if the
does
10, why
is divided
AG
arc
ference
circum=
?
GB
regular inscribed
gon
polyare
by
joined to the mid-points of the arcs subtended
tl%e sides of the polygon, the joining lines ivill form
a
of sides.
regular inscribed polygon of double tlie number
519.
vertices
the
If
Cor.
inscribed
equilateralpolygon
Ex.
962.
An
Ex.
963.
An
a
at
to
receive
regular.
about
a
circle is
Although
owing
He
nobleman.
Friedrich
he
Gauss
the
was
son
(1777-1855) was
of
a
bricklayer,he
born
was
ucation,
liberal ed-
a
to the
of his unusual
Karl
Note.
Historical
Caroline
circle is
,
Brunswick, Germany.
enabled
a
circumscribed
equiangular polygon
regular.
520.
in
tion
recogni-
talents by
to the
sent
was
a
the age
College but, at
both
it was
admitted
fifteen,
by professorsand pupils that
all that
Gauss
already knew
of
they
could
became
a
him.
teach
student
in
He
the
Unir
versityof Gottingen and
did some
important work
while
there
on
the theory of numbers.
On
his return
Brunswick,
a
humbly
private
he
1807, when
tutor, until
he
lived
astronomy
did
and
professor
in
were
analyticsteps by
there
is'no hint
as
of the
he
in
astronomy,
He
there he
also invented
B. Morse.
unusually clear,and
which
While
observatory at Gottingen
physics as well
telegraph independently of S. F.
lectures
Gauss
of
director of the
important work
His
the
as
appointed
was
the
to
developed
processes by which
he
is said to have
his proofs ; while
he
discovered
given
in his
in them
writings
his results.
248
PLANE
GEOMETRY
Proposition
521.
To inscribe
Given
circle 0.
To
inscribe
I. The
a
a
Problem
II.
in
square
given circle.
a
in circle 0.
square
construction
is left
as
II.
exercise
an
for the student.
Proof
Argument
1.
AB
_L
2.
.-.
ZAOC
3.
.-.
AC
4.
.-.
the circumference
Reasons
CD.
=
=
90".
90",
i.e.
fourth
one
By
2.
"71.
cons.
cumference.
cir3. "358.
of the
is divided
1.
into four
4.
Arg.
polygon ACBD, formed by joiningthe
pointsof division,is a square, q.e.d.
5.
"517,
3.
equal parts.
5.
.-.
III.
522.
equal
The
Cor.
discussion
The
is left
side
of
a
as
exercise
an
multiplied by V@
about
circle is equal
a
to the radius
circumscribed
Ex.
964.
Inscribe
a
Ex.
965.
Inscribe
in
Ex.
966
Circumscribe
regular octagon
a
circle
a
a
square
for the student.
inscribed
square
in
a
; the
a
in
side
to twice
a
circle
of a
square
the radius.
circle.
regularpolygon
about
a.
circle.
of sixteen
sides.
is
250
PLANE
524.
Cor.
A
I.
joining
the
regular
inscribed
525.
inscribed
vertices
of
triangle is formed by
a
r
hexagon.
A
side
of a regular
triangle is equal to
plied
of the circle multi-
II.
inscribed
the
regular
alternate
Cor.
GEOMETRY
radius
by VJ*.
Hint.
A ABD
is
is 2 B
hypotenuse
a
and
Ex.
971.
Inscribe
Ex.
972.
Divide
inscribed
in
righttrianglewhose
side B.
one
regular dodecagon
a
given circle
a
is live times
segment
one
into two
Ex.
973.
Circumscribe
an
Ex.
974.
Circumscribe
a
Ex.
975.
On
a
given line
as
one
Ex.
976.
On
a
given
line
as
one
Ex.
977.
If
radius
whose
a
is B,
is the
then
Proposition
526.
To inscribe
a
such
segments
angle
inscribed
a
in the
a
circle.
side,construct
a
regularhexagon.
side, construct
a
regular dodecagon.
inscribed
VS.
IV.
Problem
regular decagon
in
a
4?
circle.
j*
i
"
R
x
Fig.
Given
To
circle
inscribe
Fig.
1.
0.
in circle
0
a
angle
any
other.
circle.
about
regular dodecagon
"
that
a
regularhexagon
B\2
=
circle.
a
equilateraltriangleabout
side of
a
an
in
regulardecagon.
2.
in
a
circle
I.
1. Divide
and
ratio.
mean
(See Fig. 2.)
" 465.
2.
Construction
R, of circle 0, in extreme
radius
a
251
V
BOOK
circle
In
draw
O
chord
a
AB, equal to s, the
greater segment
of R.
3.
is
AS
side of the
a
required decagon.
II.
Proof
Reasons
Argument
radius
1
3. Draw
OB
4.
Then
OA
:
OM
=
OM
:
MA.
4.
"54,15.
"54,14.
"54,15.
By cons.
5.
.-.
OA
:
AB
=
AB
:
MA.
5.
"309.
6.
Also
in A
6.
By
7.
.'.A
0.45
7.
"428.
8.
\
8.
"94.
9.
10.
By cons.
"54,1.
11.
"111.
12.
13.
"424,1.
" 54, 2.
14.
"111.
=
15.
"204.
=
16.
"309.
17.
" 54, 8
1. Draw
On
2.
"
lay off
OA
A
and
2.
s.
=
3.
BM.
if^tf, Za
and
A
BM
=
10.
.-.
BM
=
11.
.-.
12.
But
13.
.:
ZMBO
14.
.-.
Z
15.
In
MAB
is isosceles,
s
=
OM.
03f.
is isosceles and ZMB0
BOM
Zo.
=
Z^Bif=Zo.
MAB
A
iden.
^45.
=
^LB
Za.
"
MAB.
A
is isosceles,
But
A
OM
OAB
~
OAB
and
9.
OA.
+
Z ABM,
=
2 Z 0.
ABO,
ZABO
+
2 Z
Z ABO,
or
-f ZMAB
Zo.
=2
+
Z0
180".
1G.
.-.
2 Zo
0
Z 0,
+
5 Z 0,
or
180".
17.,.-.Zo
18.
.-.
19.
.*.
36".
=
^LB
36", i.e.
=
the
circumference
into ten
20.
.*.
may
of the
be
18. "358.
cumference.
cir-
18.
divided
19.
Arg.
ing
by joinregular
20.
" 517, a.
equal parts.
polygon ABCD
the
^
tenth
one
a.
"
"
formed
"
,
is a
points of division,
decagon.
'jnscribed
q.e.d.
252
GEOMETRY
PLANE
The
discussion is left
Cor.
I.
III.
527.
exercise
an
for the student.
A
regular pentagon is formed by joining
vertices of a regular inscribed decagon.^
the alternate
Ex.
978.
Construct
Ex.
979.
The
Ex.
980.
Construct
Ex.
981.
Divide
982.
The
Ex.
as
diagonals of
a
of 20 sides.
regular inscribed polygon
a
a
regular inscribed
equal.
are
36" ; of 72".
angle of
an
pentagon
five
rightangle into
eightdiagonals of
angle at that vertex
a
equal parts.
regular decagon
from
drawn
any
divide
the
Ex.
983.
Circumscribe
a
regular pentagon
Ex.
984.
Circumscribe
a
regulardecagon
Ex.
985.
On
a
given line
as
one
side,construct
a
regular pentagon.
Ex.
986.
On
a
given
line
as
one
side,construct
a
regular decagon.
Ex.
987.
The
side
of
vertex
( V5
|B
1), where
"
Hint.
By
B
cons.,
regular
a
is the radius
B
: s
s
=
:
B
Proposition
528.
To inscribe
Given
circle O.
a
into
"
eightequal angles.
about
about
inscribed
a
a
circle.
circle.
decagon
is
equal
to
of the circle.
s.
V.
Solve
this
proportion for
s.
Problem
regular pentedecagon
in
a
circle,
it
To
inscribe
in circle
0
a
I.
1. Prom
equal to
A, any
a
side of
regularpentedecagon.
Construction
point in the circumference,lay off chord
" 523.
a regularinscribed hexagon.
AK
2.
lay off
Also
decagon. "
3.
Draw
4.
FE
chord
side of
a
regularinscribed
a
526.
chord
is
equal to
AF
253
V
BOOK
FE.
side of the
a
requiredpentedecagon.
II.
Proof
Reasons
Argument
Arc
AE
Arc
AF
y1^of
FE
.-. arc
the
.*.
of the circumference.
i
=
the circumference.
i
=
By
By
jL, i.e. ^
"
circumference
may
of
the
be
divided
cumference."
cir-
cons.
cons.
54, 3.
4.
Arg.
3.
5.
" 517, a.
into fifteen
.*.
equal parts.
the polygon formed
by joining the
points of division will be a regular
inscribed pentedecagon.
q.e.d.
III.
The
529.
divided
discussion
is left
It has
Note.
into the
number
been
now
of
2,
8,
3, 6, 12,
5, 10, 20,
...
-.
988.
Construct
Ex.
989.
Circumscribe
Ex.
990.
On
a
an
given
angle of
a
...
for the student.
circumference
a
indicated
16,.-.
24,
40,
15, 30, 60, 120,
Ex.
that
shown
equal parts
4,
exercise
an
as
below
as
be
:
2"^
3x2"
[n being any
integer].
5x2"
15
x
2"
positive
J
24".
regularpentedecagon
line
can
one
about
side, construct
a
a
given circle.
regular pente-
decagon!
Ex.
Assuming that it
regularpolygon of 17 sides, show
polygon of 51 sides.
Ex.
991.
992.
If
is
how
possible to
it is
inscribe
possibleto
in
inscribe
circle
a
a
a
regular
the tangents
regular polygon is inscribed in a circle,
drawn
at the mid-points of the arcs
scribed
subtended
by the sides of the insides
polygon form a circumscribed
are
regular polygon whose
vertices lie on
parallelto the sides of the inscribed polygon, and whose
the
a
prolongations
polygon.
of
the
radii
drawn
to
the
vertices
of
the
inscribed
254
GEOMETRY
PLANE
Proposition
530.
A
polygon
;
circle
may
and
circle may
a
VI.
be circumscribed
To
(a)that
(b)that
:
(a)
1.
Pass
Z
ABC
Z3
Also
.*.
.*.
OA
AB
In
=
Z2.
=
Z
=
Z4.
=
CD.
=
AOCD.
AABO
=
be circumscribed
be inscribed
like
the
polygon;
the
of the
circle,
polygon.
ABC
vertices
the
circumscribed
ABC
D.
it may
manner
of
of
circumference
through
circumference
each
through points
BCD.
and
OD
passes
10.
.
OC.
=
Z1
.-.
8.
9.
OB
.\
But
center
all the vertices
Then
6.
7.
"
about
in it.
1.
" 324.
2.
"54,15.
C.
O, the
Connect
4.
5.
"
Keasons
circumference
a
with
3.
a
regular
in it.
Argument
A, B, and
2.
"
circle may
circle may
a
any
^C
"
regularpolygon ABCD
prove
about
also be inscribed
B^-~
Given
Theorem
circle
about
proved that
passes through
of the regular
be
will
the
then
be
polygon.
Q.E.D.
it ;
BOOK
255
V
Argument
(P)
Reasons
Again AB, BC, CD, etc.,the sides
the given polygon, are
chords
1.
the circumscribed
2.
Hence
Js from
to these
3.
.*.
with
0
equal
of the circle
to
of
one
circle may
the sides
with
these
the
radius
a
Js, as
be described
of
2.
" 307.
3.
" 314.
4.
"317.
equal.
are
center, and
as
of
circle.
the center
chords
1. " 281.
of
OH,
to which
all
will
polygon
a
be
tangent.
4.
.*.
this
circle will
inscribed
be
in the
polygon.
531.
center
q.e.d.
The
Def.
of the circumscribed
532.
The
Def.
533.
of the inscribed
534.
In
Def.
angle between
of any
535.
Cor.
I.
of
as
common
0, Prop. VI.
the radius
of
the radius
regularpolygon is
a
OH.
radii of the
side,as Z
circles ;
regularpolygon is
a
regular polygon
a
is the
OA.
apothem
as
circle,
inscribed
and
as
circle,
The
Def.
regularpolygon
a
of
radius
the circumscribed
the
of
center
the
angle
polygon drawn
at
the
center
is
to the extremities
AOF.
The
angle
right angles divided
at
the
the
by
is
center
number
equal
of sides
to
four
the
of
polygon.
536.
Cor.
An
II.
supplement of
Ex.
993.
Find
regular octagon.
Ex.
994.
inscribed
Ex.
in it
995.
center is
30" ?
the
How
angle
the number
Find
If the
are
angle of
the
circle
number
regular polygon
is the
center.
degrees in the angle at the center of a
of degrees in an angle of the octagon.
circumscribed
concentric, the
many
the
at
of
a
triangleis
sides has
a
triangleand
equilateral.
about
a
regularpolygon
whose
the
angle
circle
at
the
256
PLANE
GEOMETRY
Proposition
polygons of
Regular
537.
VII.
the
Theorem
same
number
of sides
are
similar.
C
To
of
A'b'c'd'e',
polygon ABODE
prove
polygon A'b'c'd'e'.
~
Reasons
Argument
1.
Let
n
polygon; then
polygon equals
each
"
of sides of
represent the number
each
the
of sides.
number
same
and
regular polygons,ABODE
two
Given
each
angle
"217.
1.
of
A
(w-2)2rt.
n
2.
polygons
the
.*.
3.
AB
=
BC
4.
A'B'
=
B'C'
~
.
CD
=
=
2. Arg. 1,
angular.
mutually equi-
are
=
"
"
"
.
C'D'
=
"
"
"
.
j^_JW__C]?_
3.
"515.
4.
"515.
5.
" 54, 8
6.
"419.
A'B'~ B'C'~ C'D'~
6.
.'.
polygon ABODE
~
polygon A'b'c'd'e'.
a.
Q.E.D.
inches
and
of their
areas
5
homologous
sides
inches,respectively;
what
of
two
is the
3
are
regular pentagons
ratio of their perimeters ?
?
perimeters of two regularhexagons
inches, respectively; what is the ratio of their areas
Ex.
72
Two
996.
Ex.
997.
The
are
?
30 inches
and
258
PLANE
MEASUREMENT
OF
GEOMETRY
THE
CIRCUMFERENCE
THE
The
540.
straight line,i.e. its length, is
it a straightline taken as a standard
a
obtained
by laying off upon
unit (" 335).
or
Since a straight line cannot
curve,
it is obvious
must
be
that
be
made
other
some
with
coincide
to
of
system
adopted for the circumference.
will develop the principlesupon
which
theorems
OF
CIRCLE
of
measure
AND
a
measurement
The
following
such
ment
measure-
is based.
Proposition
The
I.
541.
in
inscribed
and
circle
of
the
Theorem
of a regular polygon
than the perimeare
less,respectively,
ter
regular inscribed polygon of twice as
area
sides.
many
The
II.
and
perimeter
of twice
The
Ex.
whose
(")
(6)
as
proof is
Hint.
cumscribe
regular polygon circircle are
than
a
greater, respectively,
area
of the regular circumscribed
gon
poly-
perimeter and
about
the
and
perimeter
a
area
IX.
The
1000.
radius
sum
A
as
of two
square
is 10 inches
a
sides.
many
left
of
area
exercise
an
sides of
and
;
The
difference
The
difference between
between
triangleis greater
regular octagon
a
find
a
for the student.
:
their
perimeters.
their
areas.
are
than
the third side.
inscribed
in
a
circle
542.
circumference
He
this text.
of the circle by
and
area
was
born
(285?-212 b.c.)
Archimedes
Note.
Historical
259
V
BOOK
similar to that
method
a
in
studied
Syracuse, Sicily,but
in
found
the
given in
at
Egypt
the
University of Alexandria.
garded
Although, like Plato, he repracticalapplicationsof
mathematics
yet,
his
on
Sicilyhe is said
portance,
im-
minor
of
as
to
return
have
to
won
King Hiero
by applying his extraordinary lalS
mechanical
struction
genius to the conwith
of war-engines
which
great havoc was wrought
of
admiration
the
Roman
the
on
of
means
he
focused
and
rays
rors
mir-
set. the
shipson fire. Although
Roman
be
this story may
such
untrue,
drain
so
large that
The
a
work
giveshim
could
Hiero
fixed
most
rank
and
Archimedes
was
story is told
it launched
was
used
was
A
Nile.
of the
inundations
get
which
screw,
greatest mathematicians
of the mechanics
The
Archimedes
not
the
among
circumference
impossible.
means
no
in
Egypt
ship which
moved
by
a
was
of
system
that
in this connection
by Archimedes, who remarked
itself.
the world
fulcrum, he could move
himself,however, and that which
prized by Archimedes
cogwheels devised
but
the
by
fields after the
the
to
he
be
invented
Archimedes
theless
never-
feat would
a
Archimedes
had
and
to have
is said
sun's
the
By
army.
large lenses
^W"1"?V' ^"l
area
of solids and
circle,and
of the
he
was
fluids,his
his work
all time, is his
in solid
vestigati
in-
of the
measurement
geometry
captured by the
Syracuse
drawing diagrams in the sand, as
killed when
that
of
was
Romans.
was
the
He
him.
the Roman
soldiers came
days, when
upon
he was,
but they, not knowing who
begged them not to destroy his circles,
him
with their
and thinking that he presumed to command
them, killed
his genius and
admired
The Romans, directed by Marcellus,who
spears.
to his
had given orders that he should
be spared, erected a monument
which
were
on
engraved a sphere inscribed in a cylinder.
memory,
told
is delightfully
in 75 b.c.
The story of the re-discoveryof this tomb
custom
in
those
by Cicero,who
Archimedes
found
is
it covered
regarded as
with
rubbish,
when
visitingSyracuse.
the greatest mathematician
knowh, with the sole exception of Newton.
the
world
has
260
PLANE
GEOMETRY
Proposition
543.
X.
repeatedly doubling
a
regular polygon inscribed in
polygons always regular:
I.
by
apothem
less than
The
be
can
made
the radius
of sides
making
the
differ from
to
of
the
dius
ra-
value.
of the apothem
of
number
circle, and
a
assigned
any
square
the square
from
the
By
The
II.
Theorem
by
be made
can
less than
to
any
differ
assigned
value.
Given
AB
the
side,OC
the
regular polygon inscribed
To
of the
I.
II.
OB
the radius
OB
OC
"
OB2
"
can
OC2
can
be made
be made
the
number
less than
less than
any
assigned value.
Reasons
sides
the
number
of
"519.
polygon and
making the polygons always regular,
AB, subtended
by one side AB of the
polygon, can be made less than any
assigned arc, however small.
previously
2.
,\
chord
of
the
AB
can
inscribed
be
made
less than
assignedline segment,
previously
small.
a
of sides
assigned value.
any
Argument
By repeatedlydoubling
of
AMB.
polygon :
I.
1.
in circle
and
by repeatedlydoubling
that
prove
apothem,
any
ever
how-
2.
" 301.
V
BOOK
261
Argument
CB,
which
than
any
3,
,
is
But
5.
.-.
OC
"
be
CB.
"
less
made
than
than
CB,
previously
any
small.
however
value,
assigned
less
always
being
OC,
"
can
value,
assigned
small.
OB
OB
less
made
be
can
previously
however
4.
^ AB,
Q.E.D.
II.
1.
Again,
2.
But
OC2
OB2
"
CB
be
can
CB2.
=
CB
.*.
value,
4.
0~B"
.".
being
OC
"
,
be
can
than
assigned value,
I, Arg.
viously
pre-
3.
small.
however
ously
previ-
any
3.
" 545.
4.
"309.
small.
equal
always
less
made
any
however
value,
assigned
than
than
less
made
be
can
less
made
assigned
3.
"447.
to
CB
,
previously
any
small.
however
Q.E.D.
If
544.
the
variable
a
of
quotient
less than
made
If
545.
the
value,
(For
Ex.
proofs
each
line
4
variable
of
these
the
Construct
1001.
making
each
that
line
times
be
can
12
times
as
long
as
:
except
constant,
any
assigned
any
zero,
value,
can
be
value.
assigned
variable
a
by
than
less
made
variable
any
of
square
the
be
can
less
made
be
can
theorems
as
in
than
less
(1)
designs:
the
figure
assigned value,
any
Appendix,
see
following
long
made
than
;
(2)
assigned
any
""
586
and
on
the
blackboard,
on
paper,
589.)
making
262
PLANE
GEOMETRY
Proposition
546.
circumscribed
number
regular
I.
Theorem
the
By repeatedly doubling
regular
same
XI.
number
inscribed
and
of sides, and
making
the
of sides of
polygons vf the
polygons always
:
TJveir perimeters
II. Their
areas
approach
approach
a
limit.
common
a
limit.
common
R and
K
r the apothems, and
perimeters,
of regular circumscribed
and
k the
and
areas
respectively
of sides.
inscribed polygons of the same
number
that by repeatedlydoubling the number
of sides
To
prove
of the polygons,and making the polygons always regular:
I. P and p approach a common
limit.
II. K and k approach a common
limit.
Given
P
and
p
I.
1.
the
Argument
Since the two
same
number
Reasons
regularpolygonshave
of
P_R
the
"538.
2.
" 399.
-
sides,
r
p
R
"p
"
r
_
P"p
L
"54,
P
=
7
a.
K
But
by repeatedlydoubling the number
of sides of the polygons,and making
them
be
r
can
always regular,R
"
" 543, I.
263
V
BOOK
Reasons
Argument
than
less
made
signed
previouslyas-
any
small.
value,however
~~r
5.
less than
be made
can
.-.
pre-
any
5.
" 544.
G.
" 547.
7.
" 309.
8.
" 548.
R
small.
viouslyassignedvalue,however
~r
6.
P
.-.
less than
made
be
can
any
R
however
previouslyassigned value,
small,P being
decreasingvariable.
a
D
7
being always equal to
P"p,
-\
n.
P
,
R
any previously
small.
assigned value, however
8.
andp approach a
P
.\
less than
be made
can
limit.
common
Q.E.D.
Since
Hint.
547.
the
If
exercise for the student.
an
regularpolygons
two
The
a
rest
of
have
the
proof is similar
the
548.
of sides,
number
same
2-8, " 546, I.
to steps
have
the
the less
the two
a
may
a
assignedvalue,
any
decreasingvalue
variables
such
are
that
continuallyincreases,so
be made
as
small
as
ive
these
549.
The
Note.
that
polygon
that
the
is the
theorems
above
see
the
be made
may
is
one
always
Appendix, "" 587 and 594.)
proof is limited
by
the tivo variables
them.
to
regular polygons, but
limit of the perimeter of any
same
differencebetween
please,then
limit ivhich lies between
common
shown
and
less than
other,arid if the greater continuallydecreases
(For proofs of
be
variable
related
tioo
greater than
while
be made
assignedvalue.
any
If
can
variable
product of that
less than
is
as
r2
k
can
the
El (" 539).
E-
II is left
proof of
The
II.
whatever
successively
increased,provided that
each
method
side
inscribed
the
scribed)
(or circumof its sides
number
approaches
it
zero
as
a
limit.
264
PLANE
550.
The
Def.
limit which
length
GEOMETRY
of
is the
circumference
a
common
the successive
scribed
perimetersof inscribed and circumregularpolygons (of3, 4,5,etc.,sides)
approach as the
of sides
number
zero
The
as
increased
successively
is
frequently used
(See Prop. XII.)
circumference."
a
The
lengthof an arc of a
length of the circumference
551.
of the
the
intercepts
552.
side
is of 360".
arc
as
the
for
is such
circumference
proaches
ap-
central
"the
a
par"
angle which
(See " 360.)
approximate lengthof
The
each
and
limit.
a
"circumference"
term
length of
is
a
circumference
is found
in
elementarygeometry by computing the perimetersof a series
tained
of regularinscribed and circumscribed
polygons which are obThe
of their sides.
by repeatedlydoubling the number
perimeters of these inscribed and circumscribed
polygons,
since they approach a common
to agree to
limit,may be made
decimal
as
according to the number
placesas we please,
many
of times
Proposition
553.
The
diameter
any two
Given
radii
To
R
and
prove
XII.
for
same
circles with
"_
c_=
=
-
"
,
2R
Theorem
2r'
a
circle to its
all circles.
circumferences
R',respectively.
""
polygons.
of the circumference of
ratio
is the
of sides of the
the number
double
we
Cand
(f,and
with
266
PLANE
GEOMETRY
Proposition
The
of
half the product of
557.
area
a
its
XIII.
Theorem
regular polygon is equal to
perimeter and its apothem.-
one
C
regularpolygon ABCD
Given
P
its
,
perimeter,and
r
its
apothem.
To
of
area
prove
ABCD
\Pr.
"
Argument
1. In
polygon ABCD
Then
r, the apothem
2.
"
ABCD
"
"
"
Reasons
inscribe
"
of
circle.
regularpolygon
is the radius
"
a
,
of circle
1.
" 530.
2.
" 533.
3.
"492.
0.
,
3.
of
.-. area
Ex.
1005.
Find
Ex.
1006.
The
is
mean
a
inscribed
Ex.
drawn
the
whose
regularhexagon
a
inscribed
an
figure represents
scale of 1 inch
feet in the
The
Def.
successive
as
side
20
flower
of
areas
each
to
of
area
polygons approach
and
of
area
of
area
The
of square
558.
the
q.e.d.
side is 6 inches.
agon
regularhex-
proportionalbetween the areas of the
circumscribed
equilateraltriangles.
1007.
to
number
the
and
...""jPr,
ABCD
the
a
a
flower
bed
Find
the
feet.
bed.
circle
is the
inscribed
and
number
of
approacheszero
limit which
common
circumscribed
sides
as
a
is
regular
creased
insuccessively
limit.
BOOK
Proposition
The
559.
of
its
To
XIY.
Theorem
of a circle is equal to one
circumference and its radius.
half
area
radius
circle 0, with
Given
267
V
R, circumference
C, and
Call its
polygon.
K.
regular
a
1.
"517,6.
2.
" 557.
3.
"550.
4.
" 561.
5.
" 558.
6.
Arg.
7.
" 355.
and
perimeter P
^pr.
"
the
O
S.
area
S
circle
about
Circumscribe
As
area
Reasons
Argument
Then
uct
prod-
K=\CR.
prove
its
the
of sides of the
regular
repeatedly
number
circumscribed
polygon is
doubled,P approaches C as a limit.
.*. i PR
approaches\ CR as a limit.
limit.
Also s approaches Zasa
But S is always equal to J PR.
CR.
.'. K"\
Q.E.D.
560.
The
561.
TJie limit
zero, is the limit
(Proofs of
and
product of a variable
of the product of a
of
these
the variable
a
variable
multipliedby
will be found
theorems
in
is
constant
the
OMd
a
a
2.
variable.
constant, not
the constant.
Appendix,
""
590.)
562.
Cor.
Hint.
K
The
I.
=
\
C
R
of a circle
area
=
I
"
2 ttR
.
R
=
is
iri?2.
equal
to irR2.
585
and
268
PLANE
563.
Cor.
The
II.
the squares
diameters.
of
as
564.
Cor.
radii, or
their
of
area
Ex.
1008.
Find
the
area
Ex.
1009.
Find
the
area
of
If the
1010.
Find
1011.
Find
1012.
is
area
of two
circles with
2B, AC
point on AB.
+
DB
arc
and
similar
the
\ AB, and
=
AC
AE
+
arc
arc
In
+
arcs,
are
1014.
Similar
arcs
Ex.
1015.
Similar
sectors
1016.
Similar
Hint.
sector
each
to
are
as
Prove
A AOB
AOB
_
sector
segother
of their radii.
squares
CPD
apply "" 396
segment
that
of
another, and
of the second
between
6 inches
the
and
?
ences
circumfer-
8 inches.
areas
AB
is any
;
arc
pare
(2) ComAB.
circles
different
radii.
the
CB
arc
EB.
Ex.
are
(a) 45",
similar
arcs,
sectors,and
similar
segments
sectors,
that
respond
cor-
equal central angles.
to
ments
E
semicircumference
segments
area
of the
triangle.
radius
are
is
which
4, respectively.
arc
;
of the
sum
(1) Find
Def.
Ex.
is a" is
circle
a
figurethe diameter
AD
result with
each
565.
\AB,
=
radii
of
radius
radii 3 and
In the
1013.
=
to
=
"
ring included
of the
area
the
equal
whose
AD
their
is 3 inches.
the
sector
is the
circles whose
concentric
of two
Ex.
angle
angle of
Find
circle is four times
one
inches,what
the
radius
the
sector
a
Segment
of
area
of the first is 6
the radius
circle whose
a
of
(") 120",(c) 17",and
corresponding to (6). Hint.
Ex.
other
of
squares
whose
sector
a
radius,5 inches.
the
Ex.
the
as
to each
are
(See" 551.)
360^
Ex.
circles
of two
areas
The
III.
GEOMETRY
A
and
CPD
399.
Then
to
are
each
to
other
each
as
other
their radii.
as
the
squares
of their
BOOK
Proposition
Given
566.
of
269
XV.
unit
Problem
diameter
and
polygon of n sides, to find
regular inscribed polygon of 2n sides.
circle
Given
inscribed
To
polygon of
find
of 2
n
n
in terms
x
2.
Z
3.
Also
L
.:
5.
Now
diameter
of
is
CBF
Reasons
; draw
CF-
=
1, .50
=
=#2
C5
of
AB.
1
.
1.
"54,15.
2.
"367.
3.
"142.
4.
" 443, II.
5.
By
6.
"309.
7.
"447.
8.
"54,13.
1
CO
'
2'
.%
BF.
CK.
1
CF
and
BO
rt. Z.
a
side
s.
is the _L bisector
CF
CB2
CF
CK=
2
CK=C0
"
KO
cons.
1
--K0.
=
7.
.-.
x2
=
-
"
-JW
"
-Mtjr
of
a
of
diameter,AB the side of a regular
inscribed
sides,aud CB the side of a re'gular
sides ; denote AB
by s and CB by x.
Argument
1. Draw
the
of unit
ABF
polygon
6.
the side
inscribed
regular
a
circle
a
V
ir#
Vl-s2
/rr
Q.E.F.
270
PLANE
GEOMETRY
Proposition
Given
567.
circle
a
XVI.
of
unit
Problem
diameter
and
the side
of
a
regular circumscribed
polygon of n sides,to find the side
of a regular circumscribed
polygon of 2n sides.
circle
Given
of unit
O
circumscribed
of
a
by
find
x
half
sides,and
n
polygon
of 2
n
the
CB
CO
in terms
and
of
A
in A
.-.
4.
But
5.
And
6.
.'.
7.
1.
:
AB
=
AB
"
CB
:
Z
=
CB
AO
=
+
CB
AB
:
2.
BO.
3.
"432.
4.
" 54, 11.
5.
"446.
6.
"309.
7.
"309.
8.
"403?
-[-BO
-
"
x
: x
=
yV
:
BO.
for C", and
"HS
'BO,
s
BOA.
J50
Substituting| for ^J5,
for
side
"54,15.
"517, 6.
CB.
"
"-"/
AB
ular
reg-
the
Reasons
BOC
AC
OAB,
AC
half
a
s.
AO.
2.
3.
of
sides ; denote
Argument
Draw
side
|.
by
CB
of
polygon
regular circumscribed
|and
To
diameter,AB
-
vr
+
I
:
1.
AB
271
V
BOOK
Argument
9.
s
.
"
10.
.\
Reasons
x
=
X
=
Vs2 + 1.
x
"388.
9.
10.
1 +
V
s2 +
Solvingfor
1
x.
Q.E.F.
Given
1017.
Ex.
circumscribed
a
circle of unit
; compute
square
regularcircumscribed
To
the value
Given
To
circle
regular inscribed
XVII.
the
ference
approximate value of the circumterms
; i.e. to compute
of its diameter
in
circle
A
BCD,
with
unit diameter.
approximatelythe
of its'diameter
; i.e. to
circumference
of circle ABCD
compute the value
ratio
of
tne
of
ir.
Reasons
circumference
is the
same
of
a
1.
"553.
2.
"522.
for
all circles.
the diameter
unity,the
I
and
a
the
circle to its diameter
Since
and
Problem
Argument
The
inscribed
of ir.
compute
in terms
an
octagon
compute
of a
and
the side of the
Proposition
568.
diameter
will be
side of
| V2.
of the
an
given circle
is
inscribed square
272
PLANE
GEOMETRY
Argument
3.
the formula
By using
the sides of
of
\^"-VI-
=
8, 16, 32, etc.,sides may
of the
length
polygon may
Likewise
be
if the
circle is
obtained.
square
diameter
sides
The
side of
given
will be 1.
foimula
x
=
Vi^TT
; and
of
computed
one
by multiplyingthe
side by the number
of
sides,the length
of the
polygon may
are
"567.
scribed
circum-
a
be
results
5.
sults
re-
polygons of
each
"522.
each
regular circumscribed
8, 16, 32, etc.,sides may
length of
4.
puted
com-
of
of the
1 +
the
"566.
the table below.
unity,the
By usingthe
be
perimeter
given in
are
3
by multiplyingthe length
of sides,
by the number
side
one
the
4.
x
regularinscribed polygons
; and
of
Reasons
given in
perimeter of
be obtained.
the
The
followingtable.
274
PLANE
Ex.
The
Q.
2
1022.
of wheel
of wheel
drive wheel
Ex.
is
inches.
lutions
revoolutions
rev-
many
radius
radius
The
subtended
1025.
of
a
circle is 6 inches.
of
a
circle is 8 inches.
The
the side of
by
diagonals of
of the circle inscribed
a
inscribed
an
rhombus
1026.
the number
Ex.
An
of sides
number
Find
1029.
of the
form
1030.
Ex.
angle
Ex.
in
a
square
lines
a
of
is the
be
the
of the
; in
in
area
circle is
a
regular if
about
circle is
a
of the
terms
radius
in
an
regularhexagon.
a
Find
of
the apothems
ratio of its
the
plot of
circular grass
a
of
area
a
lar
reguto the
area
6
radius
feet,a
How
many
flower
bed
square
feet
regularhexagon
a
inscribed
in
a
circle is 24
V3.
of the circle ?
area
From
is 105".
Find
1033.
Prove
a
circle of radius
the
diameter
; take
side of the
and
=
perimeter
and
OA
of the
method
CD
and
draw
required pentagon,
required decagon.
sector
a
and
OF
(ir
sector.
=
_
dicular
perpen-
point
join its mid-
FD.
central
whose
of
regular decagon
a
diameter
; bisect
ED
6 is cut
following
the
Draw
AB
EF
and
area
that
circle is correct.
D
area
?
The
1032.
to
in
equilateraltriangleis inscribed.
an
1031.
to
and
joiningthe mid-points of
inscribinga regular pentagon
.in
30 ; find the
circumscribed
regular pentagon.
a
Within
of turf remain
What
of
area
?
equilateraltriangle
16 and
are
is the
originalpentagon.
in the form
Ex.
a
is odd.
the apothem
The
pentagon
Ex.
What
inscribed
equilateralpolygon
equilateraltriangle;
area
of
area
EXERCISES
equiangular polygon
An
if the
1028.
Ex.
is the
of sides is odd.
1027.
regular
What
in the rhombus.
MISCELLANEOUS
Ex.
0 to wheel
is 60" ?
arc
1024.
the segment
Ex.
drive wheel
will the smaller
The
segment whose
Ex.
how
belt from
a
?
1023.
Ex.
0
75
makes
per minute
make
16
Q
minute,
per
wheel
figurerepresents
diameter
feet and
If the
The
GEOMETRY
FD
will
the
side
2r2.)
BOOK
Ex.
Divide
1034.
in
angle inscribed
that
so
in the
other ;
so
Show
to leave
as
a
one
how
off the
to cut
; of
regularhexagon
a
of
corners
equilateraltriangle
an
to leave
square
regular octagon.
a
regular pentagon form
1036.
The
diagonals of
Ex.
1037.
The
diagonalsjoining alternate vertices of
Ex.
a
regularhexagon
The
1038.
If
1039.
whose
radius
is
a
third
one
of
area
product of the side of
Ex.
any
an
Ex.
form
that
such
segments
times.
seven
1035.
in
two
angle inscribed in the other ;
segment is three times an angle inscribed
is twice
segment
angle inscribed
an
Ex.
given circle into
a
one
275
V
regular inscribed
a
inscribed
an
is the side of
a
2?, then
a
=
a
square
gon
regular hexa-
octagon is equal
the
and
10
the
to
diameter.
regular pentagon inscribed
v
-5-
a
one.
original
the
large as
as
regularpentagon.
a
in
circle
a
2V5*
"
29
Ex.
times
a
The
1040.
the
of
area
square
1041.
Construct
an
Ex.
1042.
Construct
a
Ex.
1043.
Through
circumference
given
Can
dodecagon is equal to
three
of the radius.
Ex.
3 to 5.
regularinscribed
a
the
angle of 9".
regular pentagon, given one
given point
a
into
construct
parts in the
two
point lie within
given
Ex.
1044.
Transform
Ex.
1045.
Construct
a
Divide
given
a
the
line which
a
ratio
of the
diagonals.
shall divide
of 3 to 7 ; in the ratio
of
circle?
given regular octagon into
circumference
equal
a
to the
square.
of two
given
circumferences
into
sum
circumferences.
Ex.
1046.
a
circle
by concentric
equivalentparts.
four
1047.
In
Ex.
1048.
In
Ex.
1049.
If two
chords
four
circles
Ex.
given sector
a
whose
angle is
a
right angle inscribe
a
square.
the
of
sum
the
equivalentto the
Ex.
whose
Ex.
equal
radii
are
1051.
to
twice
theij radii.
B
area
and
of
a
a
circle
having
circle.
perpendicular to
are
the
four
segments
as
each
other,
diameters
is
circle.
given
The
1050.
given sector inscribe
a
Rr
of
a
ring between
respectivelyis tt(B2
The
area
of the
the
area
of the
surface
smaller
concentric
two
"
between
circle.
circumferences
B'2).
two
Find
concentric
circles is
the ratio between
PLANE
276
EXERCISES
MISCELLANEOUS
1052.
If
Ex.
1053.
If,on
the
drawn
are
so
diameter,
angle, the
semicircle
a
equal
is
internal
and
righttriangleas diameters,semicircles
triangle,and if,on the hypotenuse as
of the
two
of the
area
vertex
of the
right
included
between
the
the
crescents
given triangle.
regular inscribed
and
given point draw a secant
segments.shallbe equal.
a
external
diagonals of
that the
Show
circle divide
triangleis
half that of
circle such
a
that
its
quadrilateralinscribed in
triangleswhich are similar,two
into four
quadrilateral
the
to
any
two.
Ex.
VAB
Through
JJ?2 PAx
1057.
that
so
The
circles concentric
equivalentparts
2? the
to each
Ex.
of the
Ex.
a
a
circle is 6 feet.
it whose
BC;
prove
PTa
circumferences
that
Given
a
secant
are
divide
its
the radii of the
into three
area
a
to
tangents
a
0.
circle
a
parallel
respectively
;
prove
and
TB.
AT
proportionalbetween
mean
^
that the
two
radius
unequal lines
sum.
two
similar
construct
triangles,
a
triangleequivalent
sum.
The
to the
sum
inscribed
square
circle whose
of the side of
of the squares
an
inscribed
equilateral
triangle
of the sides of the inscribed
square
and
regularhexagon.
Prove
1064.
tangent to
What
TC.
0T=\
tangent
other
two
half their
1063.
equal
construct
circle,
parallelogram ABCD,
Show
to their
Ex.
of
cutting PT at A and B
is a mean
proportionalbetween
1062.
a
B
and
1061.
is less than
of
PB.
other
of the circle
outside
?
Given
1060.
point T,
Ex.
with
mid-point of
Ex.
point P,
radius
Given
1059.
Ex.
.
a
=
1058.
Ex.
is
GEOMETRY
passing through
of the
area
From
1056.
Ex.
of
the
a
of the
areas
to
The
1055.
Ex.
is
PLANE
regular inscribed hexagon.
the
at
of
arms
is drawn
of the
1054.
Ex.
the
to lie outside
as
sum
semicircles
a
ON
the sides of
on
equilateraltriangles are constructed
given triangle,the lines joining the vertices of the given triangleto
vertices of the oppositeequilateraltrianglesare equal.
outer
Ex.
any
a
GEOMETRY
that
diameter
the inner*
the
area
equals a
of
a
chord
circular
ring is equal
of the outer
to the
circumference
area
which
chords
It two
1065.
Ex.
of
circle
a
the
chords
are
inverselyproportional.
Construct
1066.
Ex.
between
points A
two
circle which
the
segment of
a
parallelto
and
in
B
between
the
parallellines
two
circle similar
a
cumferenc
cir-
tangent through P,
lar
simi-
given
to two
parallelsis
parallelis 2 b.
a, and
two
one
and
A
through
passes
the
the
on
sum.
between
distance
The
1067.
line
a
equivalentto their
segments and
Ex.
by
point P
common
a
segments of the chords
the
and
from
drawn
cut
are
277
V
BOOK
and
P,
distance
the
Find
radius
the
is tangent to
of
other
the
parallel.
Tangents
1068.
Ex.
of
circumference
and
circle,is
1070.
The
square,
Ex.
and
altitude
the
also the
and
AD
prolonged bisects
FE
and
CD,
draw
square.
Let
square.
Find
BE.
1074.
Ex.
in the
of
13 and
triangleare
Find
solution
circle is 3
a
15
third
the
inches,
side,and
only.)
is the
inches,what
parallelsides
the
are
If F
is the
ABCD.
a
square
A
line is drawn
mid-point
locus
the
Describe
ABCD
Let
of
length
trapezoidABCD,
a
mid-point
Let
of
when
the
locus
be
E
parallelto
of the
of P
4 inches
Ex.
and
off in such
the
Ex.
enQe* of
Find
BC,
line
the
prove
that
this
cutting
line within
the
the
always remaining
moves,
exactly,and
of
mid-point
and
BE
of
segment
the
the
prove
A
a
perimeter
1077.
a
Into
as
straightline
area
of
correctness
that PM
of
a
PN=
segment
of
PK
a
"
any
in
cuttingAB
PM
"
from
point P
in
M, BC
jV,
PL.
circle whose
'height
is
inches.
side
whose
square,
way
parallelogram,and
any
Prove
the
8V3
chord
1076.
the
in K.
AD
1075.
Ex.
be
draw
diagonal AC
in P, and
and
figures,
equilateraltriangle,
of each figure?
is 12 inches.
one
contact.
answer.
your
cut
BC
the
be
P
parallelto BE.
CD
a
of
length
the
AD.
Given
1073.
Ex.
sides of
triangle. (Give
diagonals intersect at E.
whose
area
side
If the diameter
1072.
Ex.
is the
the
of 80" ?
arc
an
third
the
on
two
Find
from
joining the points of
of the
each
inches
6
point
a
inches.
is 9
the chord
perimeter of
396 feet,what
lengths of
of the
1071.
Ex.
of
area
length of
also the
If the
1069.
Ex.
through
radius
circle whose
a
tangents and
the
drawn
are
to make
of the
what
circle be divided
is 5 inches
it into
a
long, has
regular octagon.
its
Find
corners
the
area
octagon.
numbers
with
of
arcs
ruler and
less than
compasses
15
the
can
only
?
circumfer-
278
PLANE
Ex.
Through
1078.
drawn.
are
distance
On
triangle?
Ex.
line
the
of
a
the
point
chord.
is taken
Find
circle chords
a
third of the
one
the
locus
trianglesdo the altitudes meet
outside the triangle?
Prove.
?
triangleABC
a
which
D
through
of
these
is correct.
answer
boundary
Given
1080.
chords
of
class of
what
the
end
other
the circumference
on
of these
that your
In
on
point A
a
one
to the
prove
1079.
Ex.
draw
A
from
points,and
each
GEOMETRY
and
divides
fixed
point D
triangle into
a
the
within
the
side AC
on
two
;
of
parts
equal area.
Ex.
In
lengths of AC
is
area
1082.
Ex.
sides of
The
1081.
circle whose
the
to that of the
equal
triangleABC
a
and
triangle are
a
BC
the
5 and
are
12
5, 12, 13. Find
triangle.
angle
C
is
line ;
through
from
the
draw
respectively; the
given
Given
two
straightlines
that
shall be
its
between
that
tangent
of
apothem
a
whose
regular inscribed
and
the
radius
is 4 ; CA
is
and
ABC
of B
is
and
; draw
C
four
lines.
lengths are
polygon
have
straight
same
other
of these
the
is
distances
each
cut
AD
of
BA
hypotenuse
in the
not
to both
two
radius
C,
that the
straightlines
given polygons.
The
1086.
such
radius
of two
areas
Ex.
straightline
equal.
Construct
1085.
Ex.
points A, B, and
a
line shall be
circles of
of the
A
1084.
Ex.
three
Given
1083.
radius
right angle, and
a
prolonged through A to a point D so that the length of
prolonged through A to E so that the trianglesAED
is the length of AE?
What
equal areas.
Ex.
the
in the ratio
a
mean
portional
pro-
of the similar circumscribed
polygon.
Ex.
Draw
1087.
bisect
points and
1088.
Ex.
a
A
circumference
a
given
Look
Ex.
parallelogram is
of intersection,the
D
Ex.
1090.
and
E
that
its
and
diagonals
are
sides AB
are
divided
in the
same
ratio at their
point
equal.
and
iO
that the
of
a
area
are
triangleABC
of the triangleBBC
bisected
in
is twice
triangleDEB.
1091.
have
given
triangles.
respectively. Prove
that of the
Ex.
The
two
having its sides equal
given parallelogram. Show
the given parallelogram.
cnords
chords
constructed
a
for similar
If two
1089.
through
circumference.
parallelto the diagonalsof
are
parallelto the sides of
Hint.
shall pass
which
they
Two
?
circles touch
Give
a
externally.
construction
for the
How
common
many
common
tangents.
gents
tan-
BOOK
circle
1093.
and
Ex.
this
Find
1094.
point
a
to
a
of
chord
a
the chord.
to
point within
P ; line AB
the
a
touches
rightangle.
a
trianglesuch that
divide the triangleinto
shall
the vertices
of
tangents at the extremities
unequal circles touch externallyat
B respectively. Prove
angle APB
Two
circles at A
the
the
equally inclined
are
Ex.
that
Prove
1092.
Ex.
279
V
lines
three
joining
equivalent
parts.
50" and
equal to
make
at A
Find
6.
BC
with
prolonged
The
1096.
Ex.
is inscribed
triangleABC
the angle C is equal to 60".
A
1095.
Ex.
and
Prove
plane
in the
How
1098.
so
in each
12 feet
Ex.
of
1100.
a
radius
is 15
secant
and
On
1102.
a
the
its
the
radius
Ex.
radius
Ex.
them
B.
Find
7 and
an
What
whose
as
js 10
the
inches
Two
5
are
feet
the
of
center
Find
the
circle
a
of
product
the
segment.
a
if (a)
rhombus
24
are
the
-
feet and
40
feet
spectively.
re-
(6)x
-;
triangleABC.
a
=
V"
3
between
of
circumference
a
square.
locus
is the
chords
long ;
=
of
sides
included
area
circumference
1106.
diameter
a
AB, BC, CA of an equilateraltriangleABC
each
equal to one third
BE, CF, respectively,
that the sides- of triangle
triangleDEF
; prove
AD,
inscribed
1105.
point in
same
sides
side ; draw
1104.
the
circle.
is drawn.
x
Ex.
AD.
area.
Construct
1103.
of
from
inches
21
perpendicularrespectivelyto
are
from
drawn
of the
area
secant
a
diagonals of
off segments
length of
Ex.
and
through a given point in a
trianglewith two given lines
extremities
the
to
its external
Compute
DEF
drawn
chords
two
point
a
inches
The
1101.
measure
circle
Through
whole
the
D
points A
AC
drawn, viz.
passes through
isosceles
an
respectively. Find
whose
Ex.
circle is
be
can
case
lengths of
The
1099.
the circumference
Ex.
is
plane ?
Ex.
and
tangent
a
circles intersect in the
of two
of each
lines
many
form
to
as
diameter
a
does
angle
is
trapezoidare 8 and 12, and the altitude
two
trianglesformed
by prolonging the non-
straight line joining C and
the
that
Ex.
A
angle B
it ?
meet
circumferences
Through
B.
What
The
intersect.
they
The
1097.
circle.
a
a
altitudes of the
the
parallelsides until
Ex.
of
bases
to
in
how
of
center
rightangles
cuts
at
of
certain
a
the
circle
far is it from
the
a
of
given
a
circle of
given
circumference
bisect
each
center
of the
other.
One
circle ?
?
of
280
PLANE
Ex.
Show
1107.
how
length a point O which
B in the plane. If A
line at an angle of 45"
from
jB,show
Ex.
fixed
that
Ex.
at
Ex.
straightline
7 inches
distant
which
from
the
cuts
A
and
given
17 inches
13 inches.
given
when
passes,
circle.
What
prolonged, through
is the locus of the
parallelogram inscribed
and
length
diagonals?
the
Upon
given
a
angles of which is double
the oppositeside
angle meets
from
fixed
a
point
base
The
a
mid-point
of
circle
a
length.
Find
has
What
of
the
two
the
are
triangleone
a
bisector
point P.
locus
to
in
is constructed
at the
the
feet in
15
the other.
is the
What
1111.
sides
two
base
drawn
a
a
chord
certain
A
1110.
Ex.
on
point
a
a
feet in
lengths of
lie
B
will be
of
given straightline of indefinite
equidistantfrom two given points A and
on
?
1109.
sides 20
and
variable
outside
point
of the chord
find
to
shall be
OA
A
1108.
GEOMETRY
of the
largerbase
the locus of P.
the
point
different
members
of
of
contact
of
a
tangents
of
system
centric
con-
circles ?
Ex.
of which
from
Ex.
Ex.
of
sides
The
points,the perpendicular distances
each
to
are
triangleare
a
other
6,
a,
1114.
Given
1115.
In
drawn,
Ex.
triangles;construct
two
Find
c.
the
lengths of
a
square
equivalent to
A
around
Ex.
pointed
Ex.
or
has
is 10
Find
radius.
each
B
How
much
that
the
?
1117.
Prove
star) is
equal to
1118.
AB
points P, Q
1119.
and
A
circular
a
is the
and
circles ; these
Ex.
the
to
radius
the
a
of the
area
two
A'B'
same,
garden and
namely,
tween
portion be-
square
that
one
rods.
of the
angles of
are
track
yard, (tt
=
The
has
the
(a five-
rightangles.
any
two
P', Qf respectively:
running
Which
pentagram
a
chords
of the
intersect the circumference
chords
is square.
has he ?
more
sum
B
120
prove
consists
of
that^lPtwo
joined together at the ends by semicircles.
the plot inclosed by the track is 176 yards. If
track is a quarter of a mile in length,find the cost
cent
parallelchords
feet,two
chords.
land, A
more
circle whose
a
equal
each
1116.
distance
|
3 to 2.
as
medians.
these
in
of all
sum.
Ex.
are
locus
the
lines
intersecting
two
1113.
the three
their
Find
1112.
-2fa.)
outer
of two
of the
inner
PB=A'P'
centric
con-
circle
"
P'B'.
parallelstraight portions
The
extreme
the
inside
of
length of
line of
the
seeding this plot at
282
GEOMETRY
PLANE
will be used
polygon in general.
a plane figure.
side of
b
=
base
b'
=
bases
C
=
circumference
=
diameter
=
sum
D
E
of
of
of
notation
given
p
B
=
=
trapezoid.
a
of
of
a.
a
circle.
r
=
circle.
exterior
h
I
K
I
P
=
=
altitude
of
sum
angles of
a
s
=
=
line in
=
a
plane figure.
angles of a
interior
general.
perimeter of polygon
general.
in
s
=
X
=
xa
=
x0
=
a.
radius
of
regular polygon.
of regular polygon,
apothem
the
a
polygon.
of a figurein general.
area
ing
270, the follow-
projectionof b upon
radius of circle,or
or
"
polygon.
of
in "
:
=
a
to the
In addition
570.
GEOMETRY
PLANE
OF
FORMULAS
radius
of inscribed
longer of
line ;
b +
two
circle.
segments
of
or
c).
%(a +
in
general.
angle
side of a triangleopposite an
acute angle.
side of a triangleopposite an
obtuse angle.
283
FORMULAS
DLA
REFERENCE
" 457.
ma\
2
+
"
s.
465
and
Ex.
763.
"475.
"478.
"481.
"485.
ha
=
vs(s
-
a) (i
"
b)(s-c).
"490.
a
K
=
y/s(s-a)(s-b)(i
K=l(a
Polygon.
c)r.
K=\Pr.
\/g(s-q)(s-6)(8-c)
Circle inscribed in triangle.
Circle circumscribed
about
_
t
abc
B
triangle.
Bisector
b +
+
"490.
"491.
"492.
c).
of
Ex"
837"
Ex.
838.
Ex.
841.
4Vs(s" a)(s" 6)(s" c)
angle
of
angle.
tri-
2
y/bcsis" a).
b +
Trapezoid.
Regular polygons
of
same
K=\(b
P _B
c
b')h
+
"495.
_r
"538.
'
number
of sides.
r'
P'
B'
K'
JK'2
"539.
Circles.
r'*
C
C
=
2B
C
a
=
"
2B''
2tB.
" 556.
b1'
Regular polygon.
K=lPr.
Circle.
K=\C-B.
"557.
"559.
"562.
K=ttB\
Circles.
K'
Sector.
~_
553.
B'*
"563.
D'2'
central Z
_
"564.
B*.
360=
Segment.
K=
sector
^
triangle.
Ex.
1009.
TO
APPENDIX
MAXIMA
571.
conditions, the
is called
572.
all
Of
Def.
the
geometric magnitudes
greatest is called
the
that
satisfy given
and
maximum,
the
least
minimum.*
Def.
Isoperimetric
figures
Proposition
Of
573.
which
all
these
A
Given
figures which
are
have
the
To
Draw
A
the
Altitude
.'.
574.
ABC"
A
altitude
EF.
and
ABC
A
AB
ABC
Cor.
"
altitude
"
A
is
and
Z
CAE
to
an
AE
and
oblique
AEC.
*
Prove
In
later
by
reductio
mathematics
a
Z.
Q.E.D.
sides
two
then
the
are
given
given,
sides
and
include
right angle.
Hint.
AC
EF.
maximum,
a
equal
AC
maximum.
base, AC.
same
Conversely, if
I.
triangle
rt. Z
a
is the
in
AEC.
the
have
AEC
be
CAB
and
AB
sides, that
given
two
right angle
a
with
AEC,
Z
Let
A
prove
include
and
ABC
Theorem
I.
having
triangles
sides
respectively.
a
MINIMA
AND
perimeter.
same
the
GEOMETRY
PLANE
ad
somewhat
absurdum.
broader
284
use
will
be
made
of these
terms.
if
285
APPENDIX
575.
the
Cor.
is
a
and
maximum,
parallelogramhaving
the maximum
Construct
1135.
Ex.
parallelograms having given sides,
all
rectangular
is
that
one
Of
II.
two
versely.
con-
lines of
given lengths as diagonals.
Ex.
1136.
What
1137.
Of
line from
is the minimum
a
given point to
given
a
line?
Ex.
is isosceles has
which
II.
Proposition
576.
that
Of
all
and
altitude,that
Theorem
equivalent triangles having the
is isosceles has
which
base
triangleshaving the same
the minimum
perimeter.
all
the
least
base,
same
perimeter.
nCD
C
A
equivalentA
Given
and
To
let
AB=
CF
Draw
EG
BF
AB
prove
Draw
and
BC
J_
and
CA
let
and
AC
and
BE
AEC
with
+
+
the
same
base
AC,
AE^EC.
+
BC
+
and
ABC
AE
"
meet
CF
EC
the
prolongBE
CA.
of
prolongation
to meet
GC
AB
at G.
at F.
IIAC.
Z.CBF=Z.FBG.
BF
.-.
bisects
BC
=
.'.
^
'j.AB
BG
and
AB
+
AB+BC
+ BC+CA
is X
and
CG
BG
EC
CG.
EG.
=
"
AE-\-EG.
"
AE
+
"
AE
+ EC
EC.
4- CA.
Q.E.D.
286
PLANE
Ex.
Of
1138.
has
which
all
GEOMETRY
equivalent triangles having
least perimeter is isosceles.
the
the
(Prove by
base,
same
reductio
ad
that
absur-
dum.)
Ex.
1139.
perimeter
Ex.
is
Of all
the
equivalenttriangles,
equilateral.
1140.
State and
the
prove
that
one
of Ex.
converse
has
the minimum
1139.
*
Proposition
577.
Of
all
the isosceles
III.
Theorem
isoperimetric triangles
triangle is
C
isosceles A
base
and
prove
A
same
To
Draw
the
and
ABC
A
"
A
\\AC, and
E
A
AFC
is isosceles.
draw
perimeter of
A
AFC
"
.*.
perimeterof
A
AFC
"
perimeter of
perimeter of
"
AB.
A
.:
Ex.
the
one
Ex.
maximum
exterior
A
.'.
AF
.-.
FH"
AFC
that
A
has
the
What
1142.
and
what
i.e. A
point to
a
maximum
is the
the
given
the
AF
A
AEC.
A
ABC.
and
FC.
ABC.
ABC
Of all triangleshaving
1141.
having
BH.
"
"AABC;
AEC
AEC
as
ABC.
,\
.'.
A
AEC.
from
EF
AC,
other
any
perimeteras
same
ABC
BH"
base,
same
tJis maximum.
AH
Given
the
on
area
"
A
AEC.
Q.e.d.
given perimeter and
a
a
given base,
is isosceles.
chord
maximum
minimum
circumference
line
?
that
of
can
a
be
circle ?
drawn
is the
What
from
a
given
287
APPENDIX
Ex.
Of
1143.
has the maximum
all
triangleshaving
is equilateral.
area
given perimeter, the
a
IV.
Proposition
Theorem
polygons having all
equal, respectively,to given lines
all
Of
578.
maximum
be
can
undetermined
in
inscribed
side
that
one
their
taken
in
semicircle
a
but
sides
one
order, the
the
having
diameter.
as
C
H
A
to the
that
prove
diameter
does
not
that
Z
Then
if the
a
rt.
of the
But
of all
Z,
ject
polygons sub-
equal respectively
are
in order.
semicircumference
and
described
vertex,
some
with
AH
as
F.
semicircumference
with
as
AH
Draw
E.
diameter
as
AE
and
EH.
rt. Z.
a
figuresABCE"nd
polygon ABCEFH
any
a
the
is not
AEH
becomes
.-.
the
through
pass
Then
AEH
taken
through B, C, E,
passes
Suppose
maximum
that AB, BC, CE, EF, FH,
condition
given lines
to
To
the
polygon ABCEFH,
Given
A
revolved
EFH"re
will be increased
AEH
be increased
can
in
area
about
in
E
until
area.
without
ing
chang-
given sides.
this contradicts
the
hypothesisthat polygon ABCEFH
is
maximum.
.*.
the
suppositionthat
vertex
E
is not
the semicircumference
on
is false.
.-.
the semicircumference
In the
same
polygon lies
Ex.
1144.
on
through
proved that
passes
it may
be
way
the semicircumference.
Given
"%ictthe triangleso
the
hase
that its area
and
the
every
vertex
of the
q.e.d.
vertex
shall be
E.
a
angle
maximum.
of
a
triangle,con-
288
PLANE
Ex.
Find
1145.
drawn
from
the
it to
GEOMETRY
point
given
a
in
given straightline
a
circle contain
Proposition
Of
579.
all
that
polygon
Given
polygon A'b'c'd' which
AB
A'B',BC=B'C', CD
"
To
From
any
draw
a
sides
spectively,
equal, reorder, the polygon
in
circle is
maximum.
a
is circumscribed
circumscribed
be
cannot
"D', and
=
angle.
DA
by
by
O, and
a
a
O,
with
and
ED.
D'A'.
=
A'b'"D'.
A
as
draw
equals CD,
diameter
AE;
Ad'c'e'
construct
draw
E C
equal to ADCE;
A'e'.
diameter
circle whose
The
the
"
vertex
C'D',which
On
pointsB', C',D'.
either
.*.
can
ABCD
prove
by
which
ABCD
their
have
taken
circumscribed
be
can
lines
given
to
maximum
a
that the tangents
Theorem
V.
that
polygons
such
ABCE
or
less,than
be
the
is A'E' does
not
pass
through
all
(Hyp.)
EDA
or
both
must
be
correspondingpart
greater,and
of
neither
polygon A'b'c'e'd'
(" 578).
.-.
ABCED
.'.
Ex.
ABCD
"
A'B'C'E'D'.
"
A'B'C'D'
1146.
In
a
1147.
Of
all
given
But
semicircle
A
DCE
=
A
D'C'e\
Q.E.D.
inscribe
a
trapezoidwhose
area
is
a
maximum.
Ex.
number
of
sides,the
polygons having a given side
equilateral
that is regularis a maximum.
one
and
a
given
290
GEOMETRY
PLANE
VIT.
Proposition
Of
582.
has
the
two
To
prove
In
one
EFHG
.*.
P
"
of sides
EFHG
"
any
; i.e. P
as
has
Of
the
pointas
as
Q, and
To
"
square
Construct
P
an
have
irregularpolygon having
Q.
one
S
=o=
the
Q.E.D.
VIII.
Theorem
equivalent regular polygons,that
two
greater number
prove
let
H.
of sides
has
H
Given
area.
P.
Proposition
583.
and
greater
Q.
be considered
number
P
the
has
Q.
side of Q take
may
same
of sides
polygonsP
isoperimetric
the
side than
more
isoperimetric regular polygons,that which
greater number
Given
Theorem
regularhexagon
H.
perimeterof S " perimeter of H.
with H.
square R isoperimetric
the
which
smaller
rimeter.
pe-
291
APPENDIX
of
Area
.*.
sides
Of
Cor.
and
a
is
perimeter
Ex.
1149.
that has
Ex.
of
area
all
S
"
perimeterof
R.
S
"
perimeter of
H.
that
all
AND
variable
a
a
I.
that
prove
less than
V, which
Assign
any
Now
value
a
Take
v
"
than any
586.
the
given circle,
a
v
K
"
may
for V may
Then
~.
v-
If
I.
less than
any
and
constant
small,and
be made
any
assigned
let
assigned value, the
except
zero,
i.e. V-
a;
K
any
be any
as
small
as
ously
previ-
constant.
please,i.e.
we
K
as
we
please.
be
may
less
made
q.e.d.
can
be made
quotient of the
be
can
can
small.
how
be found
K"
less than
small
as
matter
variable
a
scribed
in-
perimeter is regular.
assigned value.
Cor.
one
of sides and
number
be made
can
assigned value.
value,as a, no
any
minimum
a
Theorem
be made
assignedvalue,however
To
of
assigned value.
any
variable
a
number
THEOREMS
LIMITS.
can
less than
Given
in
maximum
value, the product of the variable
be made
q.e.d.
has
given
a
that has
one
Proposition
If
which
polygons having
VARIABLES
585.
of R.
area
given
a
trianglesinscribed
the
given circle,
a
"
perimeter is equilateral.
Of
1150.
S
polygons having
the
Among
of
R ; i.e. area
given area,
regular.
maximum
a
in
"
perimeterof
perimeterof
.-.
584.
H
made
less than
variable
less than
by
any
stant,
con-
any
assigned
any
value.
Hint.
587.
;=
"
"
A
K
Cor.
II.
"
V,
which
If
a
is the
product
variable
can
of the variable and
be made
assigned value, the product of that
a
constant.
less than
variable
and
any
a
de-
t
292
PLANE
value
creasing
GEOMETRY
be
may
less than
made
assigned
any
value.
Apply
Hint.
value
any
588.
precedingtheorem, using as
decreasingmultiplier.
of the
Cor.
be
may
the
Cor.
assigned value, the
less than
Ex.
theorem:
The
"
point within
product of
"
The
not
the
by
R
.\ K-
be made
can
Prop.
I is illustrated
drawn
chord
II.
by
through a
the
fixed
Theorem
a
variable
and
variable
multiplied
limit
of
V
which,
approachesthe
constant
the limit of
L
=
V; then
"
L
V=K-
the limit of
K-
"
the limit of
But
.-.
be any
K
prove
Let
variable
any
let
To
a
of the product of
is the
zero,
variable
any
the
a
stant,
con-
constant.
Given
and
limit
less than
?
Proposition
590.
variable
a
(Apply Cor. II.)
segments of
the
circle is constant
a
that
of
square
of the corollaries under
Which
than
greater
and
be made
can
assigned value.
any
1151.
variable
a
value
variable.
a
or
If
IV.
a
product of a variable
constant
a
589.
The
III.
K
K
K
.
not
K
"
zero.
V=
V"L
finite limit i,
"
K
"
L.
R.
R.
.
R
V=
0.
=
the limit of
(K
"
L
"
K
"
R)
=
K
"
L.
Q.E.D.
591.
a
The
Cor.
of the quotient of
limit
is the limit
constant
of
the
variable
a
variable
divided
by
by
the
constant.
Hint.
"
K
=
"
"
F, which
is the
Proposition
592.
then
If
product of
the variable and
a
constant.
jBl
two
the limit
of their limits.
variables
of their
III.
Theorem
approach finitelimits, not zero,
product is equal to the product
293
APPENDIX
variables
Given
approach the
V1 which
and
V
finite limits
L\ respectively.
and
To
the limit of
prove
Let
R
V
.-.
"
V=
L
V' =
L
"(L'
of (jJ
"
the limit of
.-.
F
=
the limit of
593.
"'
+
R
L1.
"
r'.
"
"
R!.
"
L
+
R
"
L
==
R'
"
"
R
"
i?'
R
-
-
=
"
the limit of
R^.
i?') 0.
[L
i'
"
(l'
.
"
i?
i
+
R'
"
F
"
V' =
eac/z/
7/
Cor.
"
"
of
i'.
q.e.d.
of variables
finite number
any
approaches a finite limit, not zero, then the limit
their product is equal to the product of their limits.
IV.
Proposition
If
594.
the
the
that
the
less
differencebetween
the
please, then
we
as
the two
let
AP, and
AP
that
decrease, so
zero
To
as
that
prove
lies between
which
^
AP
and
every
AP'
*j$utAP
value
assumes
is
two
be
may
variables
L
R'
and
AP
that
as
between
AP1 have
tinually
con-
made
have
a
as
mon
com-
Q' P'
AP', AP' greater than
AP
AP
increases AP' shall
and
AP'
shall
proach
ap-
limit,as AL,
a common
which
AP'.
values
correspondingvalues
Since
if the greater
limit.
and
AP
such
difference
the
successive
Denote
the
a
AP' be
is
one
them.
related variables
and
that
continually increases,
two
R
Q
P
A
Given
the
lies between
which
limit
such
are
other, and
while
decreases
of
Theorem
variables
related
two
always greater than
small
"
L-L\
R.R')]
.-.
L'
V' =
"
V'
"
=
L'
"
V
F' =
and
R
"
the limit
But
R'
and
L"V
=
Then
so
L
which
of
of AP'
by
.-.
AQ, AR, etc.,and
denote
by AQ',AR', etc.
assumes"
AP
(Hyp.)
AP
AP
"
continually
increasing.
is less than
AR'.
any
value
294
PLANE
GEOMETRY
Hence
Since
value
has
AP
which
But
Hence
is
AP1
Then
Then
has
the limit of
and
the
limit
AK1
their difference
But
is
limits
This
AP1
the
.*.
and
ap\
of
AP
of AP'
is AK*.
and
m
ra',and
d.
value
less
be
of
and
AP
the
=
the
of AP'
limit
difference
is associated
of
and
between
lies between
AL.
as
responding
cor-
the difference
than
hypothesis that the
approach zero as a limit.
AP
q.e.d.
Theorem.
595.
values,as
some
cannot
that
the
shall
limit
limit," 349.)
a
AK'.
and
AK
"
while
AK,
value,as
between
every
AR.
limit of APf.
finite
some
finite
a
be
AP
have
of AP'
contradicts
and
AP
AP'
def. of
(By
=" the
AP
of
difference
value
the two
limit.
limit," 349.).
a
is greater than
.*.
some
let the
AK
def. of
assumes
assumes
Q' P'
R'
(Hyp.)
continuallydecreasing.
AP
Suppose
L
(By
AP*
which
value
any
AP1
limit.
some
R
Q
P
A
a
With
number
straight line segment
every
which
its
be called
may
there
measure-
number.
line
For
where
case
chosen
the
"" 335
segment
is
336;
unit
shall
we
now
incommensurable
straightline segment
the
to express
Apply
in terms
a
of
this
rem
theo-
consider
with
the
u
'
apply some
and
a
the unit
segment
u;
u.
(as a measure) to a
t times, then
t.u"a"(t
Now
and
the
unit.
Given
v
with
commensurable
in
considered
was
the
segments
fractional
as
many
+
part
times
as
possible,suppose
l)u.
of
u, say
-,
to
a, and
sup-
295
APPENDIX
tx times,then
it is contained
pose
p
p
apply smaller
Then
"
-,
and
",""",
"
,
smaller
and
them
suppose
fractional
parts of
to be contained
U.
to a, say
u
ts"tu
times
"""
p2 ps p*
then
respectively,
h
-2-
^to + 1
a"^"
^
u"
-.
pi
pi
^
"
u
.
"
a
p6
infinite series
the
Now
+ l
^t3
-?J--
U
J.
u,
"
of
-u,
pi
increasing numbers
t, J,
p
of which
none
exceeds
(thelimit
n
number
of
of this
with
a
which
series)
the unit
then
m
a, and
"
u
'
if
if
justifiedin associatingthe
that
n
"
596.
u
is any
n, then
"
m
shall call the
we
a
u
"
number
"
a
measure-
that
such
is
n
"
m
n,
therefore
are
we
;
with
n
""",
parts into which
of
number
m
,
ber
num-
Moreover, this number
u.
m
1, defines
t+
(thenumber
of p
for
divided),
was
to
respect
unique, i.e. independent
.
finite number
the
"
p2
in
a, and
saying
a.
=
procedure may be applied to any
geometric magnitude has a
every
Manifestly,the above
Note.
geometric magnitude
whatever,
i.e.
unique measure-number.
597.
If
Cor.
a
limit, then,
a
the
as
measure-numbers
of
the measure-number
Discussion
598.
To
determine
their
common
every
Hence,
measure
wlwther
two
if they
is the
GD
measure
every
of
problem
GE
of
common
and
given
AB
is
ratio
their
and
a
lines
surable
commen-
are
commensurable,
greatest common
measure
CD
:
are
and
successive
approach as their limit
of the magnitude.
variable
of the
measure
Moreover,
For
the
approaches
and
varies, the
magnitude
of the limit
not; and
or
is variable
magnitude
of
of
therefore
AB
and
a
measure
find
(" 345).
of
measure
measure
to
AB
and
its
multiple
GD
is
a
CD.
GE.
common
of their differ-
296
PLANE
therefore
and
ED,
ence
Hence, every
of
measure
and
their difference
exceed
of
and
AB
of
AB
and
a
measure
and
a
multipleof
is
CD
common
no
common
and
ED
(a multipleof FB)
EG
a
ED
of their difference
of
measure
and
of
measure
is
FB
hence
a
of
and
AB
is the greatestcommon
Therefore,GD
GD.
is
common
Hence,
GD.
which
therefore
and
AF
of ED
measure
AF,
measure
Again, every
common
can
of
common
AB
FB.
GEOMETRY
CD
measure
CD.
Now, if
and
AB
; for
CD
commensurable,
are
of
measure
common
any
AB
the
and
must
process
is
CD
a
minate
ter-
measure
of each
remainder, and every segment appliedas a measure
less than the preceding remainder.
Now, if the process
not terminate,a remainder
which
could be reached
would
less than any assignedvalue,however
small,and therefore
than
If
which
is
did
be
less
is absurd.
the
greatest common
AB
minate
incommensurable, the process will not terwould
obtained
be a
for,if it did, the last remainder
and
;
CD
are
of
measure
common
and
AB
An
Theorem.
599.
measure,
CD,
angle
as
shown
above.
be bisected
can
by only
one
line.
.C
Z ABC,
Given
by
bisected
^"^^
BT.
To
that
prove
bisector
ABC
exists.
that
another
of Z
Suppose
of Z
bisector
other
no
^
j,
^
Z?-"^-^^^^^
"~
'
"
exists,
ABC
"
'A
e.g. BF.
Z
.-. no
other
bisector
600.
Note
on
and
ABF
may
magnitudes.
=
be
This
Z.ABT.
Then
of A
used
Axioms
when
2-9
impossible.
exists.
ABC
The
Axioms.
is
q.e.d.
bers
(" 54) refer to numof geometric
measure-numbers
thirteen
axioms
referringto the
not
are
applicable always
to
equal figures.
8 hold for positivenumbers
7 and
(See Exs. 800 and 801.) Axioms
for infinity
;
only, but do not hold for negative numbers, for zero, nor
axioms
11 and
12 hold
only when
the
number
of parts is finite.
298
GEOMETRY
PLANE
parallel
the
locus
The
(b)
to
the
of
is the
line
given
a
of
mid-points
chords
all
of
a
circle
perpendicular
diameter
line.
to
'
R
f(-
"~
(
"
a\j""
To
0, line
circle
Given
circle
that
0
and
AB,
locus
the
RS
prove
~rp
diameter
of
the
RS
_L AB.
mid-points
chords
of
all
8.
" 130
II AB.
are
Argument
Let
be
P
Through
Now
RS
.'.
RS
.-.
P
_L
Q
Then
.-.
T
.'.
RS
II AB.
FG
be
of FG,
mid-point
point
any
in
T.
AS
J_
is the
the
RS.
FG.
Through
RS
diameter
J. AB.
is the
Let
draw
P
in
point
any
draw
Q
IIAB,
II K
IIAB.
diameter
in
not
chord
RS.
intersecting
HK.
mid-point
mid-point
is the
all chords
a
locus
of
of
HK,
of
of circle
HK,
a
the
0
i.e. Q
chord
is not
II AB.
mid-points
that
are
of
II AB.
Q.E.D.
of
GEOMETRY
SOLID
BOOK
PLANES,
LINES,
602.
Solid
Def.
VI
AND
geometry
ANGLES
the geometry
or
figureswhose
parts are not all in the same
of plane or plane surface,see " 34.)
the definition
From
603.
of
a
of
treats
of space
tion
plane. (For defini-
plane it follows
line
(a) If two points of a straight
lies in that
SPACE
IN
that
plane, the
lie in
a
plane
in not
:
whole
line
plane.
line
(b) A straight
intersect
can
a
than
more
one
point.
plane is
(length and breadth)
only a portionof it can
Since
604.
in
shown
be
is
This
a
a
unlimited
in
its
two
dimensions
figure.
usuallyrepresented
by
a
drawn
parallelogram.
rilateral
quadas
a
Thus
represents a plane. Sometimes, however, conditions make
than
to represent a plane by a figure other
it necessary
MN
as
parallelogram,
Ex.
in
the
a
1152.
vertical
Draw
a
in " 617.
a
rectangle freehand
in
plane ; (b)
rectangleof (a) be drawn
a
299
is
supposed
to lie :
(a)
angles of
plane. May
those of the rectangleof (5) ?
horizontal
equal ?
which
the
four
300
SOLID
605.
In
Note.
the
GEOMETRY
figuresin
solid geometry
represent all auxiliarylines and
to
which, for
but
lines will be
it
continuous.
In
for
great help,
a
a
represent
pasteboard
lines perpendicular to
and
time
at
with
practicallyno
expense.
For
by high school students,see
633, 678, 756, 762, 770, 797.
revolve
about
contain
can
line in it
a
as
an
particularpoint
any
in
dent
stu-
he
so,
may
figures. By using
or
small
to
sent
repre-
expenditure
reproductions of
p. 302
on
it does
models
also ""
;
A
postulate.
axis,and
the
represent oblique lines,
group
Revolution
20.
Assumption
If
stiffwires
comparatively
a
actuallymade
606.
the
visible
All other
figure.
figures.
make
stringsto
with
to be
in solid geometry
sticks of wood
plane, and
a
in the
the
imagining
least,to
actually made
be
figure may
of time
the earlier work
lines will be used
supposed
not
are
represented
planes, thin
to
any
proof,are
in
experience difficulty
may
find
of
purposes
lines that
dashed
plane
622,
may
so
revolve,it
one
and
only one
follows
that
as
in
space
position.
the revolution
From
607.
Through
a
postulateit
given straightline
number
any
:
may
be
(" 606)
it
of planes
passed.
For,
plane
as
occupy
may
unlimited
of
each
will
of
608.
Def.
axis
ber
num-
if that
a
A
plane is said
plane
From
609.
?
why
1154.
line
does
How
1153.
and
no
"" 607 and
straightline
a
an
plane
AB.
Ex.
as
which
through
in space
AB
an
represent
Ex.
about
positions
different
A
revolves
MN
not
many
to be
other
608
plane
it is
a
be
tions
condi-
fulfills those conditions.
that:
seen
determine
planes may
by given
determined
plane.
passed through
any
points
two
?
At
a
point
P
perpendicular to
in
AB.
a
given straightline
How
many
such
AB
lines
in
can
space,
be
struct
con-
drawn
?
BOOK
AND
LINES
I.
plane is determined
A
point
not
301
PLANES
Proposition
610.
/^
VI
Theorem
by
straight line
a
and
a
the line.
in
1*
^
N
A-""
--*
/_
/
^
f
'
/j'
line
Given
To
and
AB
that
prove
P,
and
AB
in
point not
a
determine
P
AB.
a
plane.
Argument
1.
Through
2.
Revolve
until
pass any plane,as
planeMN about AB as
AB
it contains
plane in
3.
Then
this
plane RS
point P.
4. Furthermore, in no
plane MN, in its
contain point P.
5.
6.
RS
is the
AB
and
P.
.'. AB
and
P
.-.
611.
in the
Cor.
same
I.
line
other
determine
A
a
plane is
the
and
AB
positioncan
rotation
only plane that
axis
an
Call
point P.
positionRS.
contains
MN.
about
can
AB,
contain
plane.
q.e.d.
determined
by
three
points
straight line.
A, J5,and C be the three given point??. Join A
straightline,and apply " 610.
Hint.
612.
Let
Cor.
not
II.
A
plane is determined
straightlines.
by
two
and
B
by
a
ting^
intersec-
302
GEOMETRY
SOLID
613.
Cor.
A
III.
plane
is determined
by
two
parallel
straight lines.
Ex.
through P,
Ex.
lines ?
Ex.
In
how
why
?
pencilsso
two
ways
can
pencilsbe held
two
If so, how
?
a
plane
that
so
the
then
measure
Ex.
1159.
Ex.
1160.
three
a
common
614.
all
common
can
be
bushel
across
measure,
passed
the
the
top.
surveyor's transit or a photographer's
three legs rather than on two or four ?
is
ure
meas-
Why
is
a
camera
many
The
Def.
pointscommon
615.
in
point ?
plane
planes are determined
by four straightlines,
lie in the same
plane, if the four lines intersect : (1) at
(2) at four different points?
How
of which
no
full ?
even
Why
always supported on
no
he
can
?
In
1158.
struct,
Con-
in AB\
AB.
measuring wheat with a half
is first heaped, then a straightedgeis drawn
Ex.
point not
a
passed through
done, assuming that the pencilsare
that
this be
P
#
Can
them
and
in space,
perpendicular to
line
many
1157.
through
a
Hold
1156.
them.
line AB
Given
1155.
intersection
to the two
Assumption
21.
also have
Reproduced
Models
surfaces
two
is the locus of
surfaces.
Two
Postulate.
another
from
of
point in
made
planeshaving one point
common.
by High
School
Students
Proposition
616.
If
-planes
two
303
VI
BOOK
IT.
Theorem
intersect, their
intersection
is
a
straight lineS
N
planes MN
intersecting
Given
To
the intersection
prove
and
of
RS.
and
MN
RS
a
str. line.
Argument
1.
Let
A
and
be any
B
to the two
Draw
both
Likewise
str. line AB
and
2.
" 54, 15.
3.
" 603,
a.
4.
" 603,
a.
AB
5.
" 610.
planes MN
6.
" 614.
7.
Arg. 2.
Args. 6
RS.
lie in
plane
plane MN,
MN.
lies in
plane RS.
outside
point
of
planes.
intersection of
is
AB
a
str. line.
the
intersection of
str.
line.
1161.
Is it
?
and
MN
possiblefor
Explain.
for.your answer.
Is the statement
than
more
1162.
1163.
RS
is
8.
a
and
7.
q.e.d.
By referringto ""
expression,"Two
planes determine
Ex.
" 615.
RS.
straightline
Ex.
no
is the
AB
B
lies in
lie in both
can
Ex.
and
A
Furthermore
.-.
and
planes MN
str. line AB
But
1.
points common
two
str. line AB.
Since
.-.
Reasons
in
26
a
Ex.
and
two
planes
608, give
the
to intersect
meaning
in
of the
straightline."
11G2
always
true
?
Give
a
reasons
304
SOLID
GEOMETRY
Proposition
617.
III.
Theorem
If three
planes, not passing through the same
line, intersect each other, their three lines of intersection
else they are
are
concurrent,
or
parallel, each to each.
M
M
Fig.
planes
Given
lines MN,
To
Fig.
PS, and
MQ,
RS-, also
and
MN
RS
RN
2.
intersecting each
intersectingat
0
concurrent.
RS
Reasons
."
0
is in line MN,
."
O
is in line RS, it lies in
".
O, lying in planes MQ
it lies in
plane MQ.
plane PS.
and
PS, must
PQ
passes
PQ.
intersection,
through 0; i.e. MN,
and
RS
concurrent
\
are
II. Given
To
IIMN
PQ
prove
in 0.
PQ,
1.
" 603, a.
2.
" 003,
3.
" 614.
4.
Arg.
2.
and
MN
Suppose
MN
are
that
either
PQ
and
RS.
But
4.
.-.
5.
Likewise
this is
PQ
Reasons
||or
II.
not
intersects
also intersects
3.
MN
" 161, a.
; then
MN
II RS.
MN.
PQ
" 617, I.
RS.
for
impossible,
3.
4.
IIRS.
3.
q.e.d.
Argument
1. PQ
a.
(Fig.2).
IIRS
MN
in
(Fig.1).
Argument
lie in their
other
:
PQ, and
MN,
prove
M
1
PQ, and
I. Given
N
Q.E.D.
5.
By hyp.
" 161, b.
By steps
to
ilar
sim1-4.
306
SOLID
*
GEOMETRY
Proposition
IV.
Theorem
If a straight line is perpendicular to each of
section,
intersecting straight lines at their point of interit is perpendicular to the plane of those lines.
622.
two
str. line FB
Given
To
AB
and
BC.
prove
FB
J_
_L
and
AB
meeting
2.
MN
MN
taining
con-
draw
of
Proof
through
AC;
J?
draw
any
draw
AF, HF, CF, AE,
line,as
BHy
at H.
AC
Prolong
FB
to E
that
SO
BE=FB',
CE.
HE,
3.
FC
plane
plane
plane MN.
Outline
1. In
at B, and
to BC
AB
and
BC
then _L bisectors
are
of
FE
i.e.
;
CE.
=
FA
=
AE,
)
4.
Prove
A
AFC
=
A EAC
; then
Z
5.
Prove
A
HAF
=
A
; then
HF
6.
.-.bhA-FE;
i.e.
EAH
FB"BH,
any
HAF
=
Z EAH.
=
##.
line in
plane 3fiV passing
through B.
7.
.'.
623.
to
a
FB"MN.
Cor.
All
the
straight line
at
plane perpendicular
perpendiculars
a
given point
that
in
to the line at the
the
can
line
be drawn
lie in
given point.
a
BOOK
Proposition
Through
624.
to
point P
To
to
construct
a
plane
pendicular
per-
1
and
line AB.
through P,
construct,
Problem
given line.
a
Fig.
Given
V.
given point
a
307
VI
plane _L
a
AB,
I. Construction
1. Through
line
and
AB
point P
a
pass
plane,as
APD
(in
Fig. 1, any plane through AB). "" 607, 610.
2. In plane APD
construct
PD, through P, J_ AB.
"" 148, 149.
3. Through AB
" 607.
pass a second plane,as ABC.
4. In plane ylPC, through the foot of PD, construct
a _L to AB
(PC in Fig. 1, DC in Fig. 2). " 148.
5.
Plane
The
II.
Hint.
jh
proof is
The
Ex.
1164.
Ex.
1165.
and
AB
the
left
by
as
C, D, and P, is the
plane required.
exercise for the student.
an
Apply " 623.
III.
Are
MN, determined
same
discussion will be
Tell how
Lines
CD
given in "
to test whether
AB
and
CD
are
necessarilyparallel?
plane ?
why
or
why
not
?
or
not
625.
a
flagpole is erect.
perpendicular to line EF.
Explain. Do they necessarilylie
each
308
SOLID
GEOMETRY
Proposition
625.
Through
perpendicular
VI.
Theorem
given point there exists only
line.
a given
a
to
plane
one
S^
^8
::::y /r"^^p/
M"
M*-
*
1
\B
\B
Fig.
'
1
1.
Fig.
plane MN, through P, _L AB.
the only plane through
MN
prove
2.
Given
To
Argument
1. Either
MN
is the
2.
In
draw
5.
Then
6.
This
7.
.*.
"
P
AB.
Only
only plane through
P
"
AB
or
it is not.
line
through P intersectingline AB, as PR.
3. Let plane determined
by AB and PR be denoted by APR.
4. Suppose that there exists another
plane through P A. AB;
let this second plane intersect plane APR
in line PS.
MN
_L
AB
also PS;
MN
627.
"" 624
Through
Cor.
from
In
625
and
630.
are
_L
AB.
Def.
the
Def.
Def.
exists
in
and
one
q.e.d.
statement:
one
only
one
plane perpendicula
given line.
a
I.
of all points in space equidistant
extremities
of a straight line segment is
The
the
locus
to the
segment
A straight line is parallel to
plane cannot
A
Two
parallelto
planes
at
its
mid-point.
a
plane if the straight
a
plane if
meet.
straight line is oblique
nor
perpendicular
631.
PS
JL PS.
be combined
may
plane perpendicular
line and
and
AB.
AB
Fig.2, explain why
given point there
a
to
629.
i.e. PR
impossible.
is the only plane through P_L
Question.
the
and
PR
is
626.
628.
a
are
the
to
it is neither
plane.
parallel if
they cannot
meet.
BOOK
VII.
Proposition
line
parallel.
are
Theorem
planes perpendicular
Two
632.
309
VI
the
to
same
straight
a
N
*
V
M
7
To
and
planes MN
Given
Use
Hint.
indirect proof.
Compare
Proposition
If a plane
633.
of intersection
and
MN
To
634.
.Ex.
are
in AB
prove
Hint.
same
RS
AB
Show
Cor.
and
187.
Theorem
parallel planes, the lines
two
parallel.
and
CD,
and
RS, and
plane PQ
respectively.
CD
Parallel
parallel planes
1166.
"
any
intersecting
II CD.
that AB
I.
line AB.
with
VIII.
intersects
IIplanes MN
Given
"
IIRS.
MN
prove
RS, each
State
the
are
converse
cannot
meet.
lines
intercepted between
equal. (Hint. Compare with
of
Prop. VIII.
Is
it true?
the
" 234.)
310
SOLID
GEOMETRY
Proposition
If
635.
IX.
angles, not
two
in
Theorem
the
plane, have
same
sides
the
the
equal-*
and
parallelrespectively,
lying on
line joining their vertices, they are
Z
and
BC
of line
To
Ex.
in
plane
IIrespectively
to ED
Given
ABC
MN
and
and
Z
in
DEF
EF, and
lying on
sidle
same
plane
their
RS
the
with
of
BA
side
same
BE.
prove
1167.
*
Z
Z
ABC=
Prove
It will also
be
Prop.
seen
BEF.
IX
if the
(" 645) that
the
angles lie
planesof
on
these
opposite sides
of BE.
angles are parallel.
If
636.
other
plane, the
perpendicular
perpendicular to the plane.
two
-parallel lines
also
is
of
one
Theorem
X.
Proposition
311
VI
BOOK
is
to
a
A
B^
^E
To
and
IICD
AB
Given
plane MN.
plane MN.
_L
CD
prove
_L
AB
Reasons
Argument
Through
1.
D
draw
any
B
draw
BE
line in
DF.
as
2.
Through
3.
Then
4.
But
5.
.'.
Z
Z
Z
ABE
is
is
CD.F
6. .\ CD
X
Z
=
XB#
line in
a
a
in
plane MN
CDF.
3.
" 635.
rt. Z.
4.
" 619.
5.
" 54, 1.
6.
"619.
rt.
plane MN
Z; i.e. CD
through
_L D.F, any
D.
Q-E-D-
planeMN.
In the
1168.
=
Prove
each
Ex.
Can
1169.
of two
CD
lie in the
same
=
a
line be
perpendicular to
intersecting
planes ?
If
1170.
one
of two
Prove.
planes is perpendicular
given line,but the
planes are not parallel.
to
is not, the
and
answer.
your
Ex.
2. " 179.
IIDF.
accompanying diagram AB
S5", angle BCD
plane. Angle CBA
35",
ABE=
in
90", .RE lying
angle
plane Jf2V. Is
CD
to
necessarilyperpendicular
plane MN?
Ex.
1. " 54, 15.
plane MN,
a
other
to
straightline and a plane are each perpendicular
"s*j,me
straightline,they are parallelto each other.
Ex.
1171.
If
a
the
812
SOLID
GEOMETRY
Proposition
637.
Through
to
XL
Problem
given point
given plane.
to
a
a
construct
a
line
pendicula
per-
R
rN
M
Fig.
pointP
Fig.
2.
and
plane JI/iVT.
construct, through P, a line
Given
To
1.
_L
plane MN.
I. Construction
In
plane MN draw any convenient line,as AB.
" 624.
Through ^construct
plane PQ J_ AB.
in CD.
Let plane PQ intersect plane MN
" 616.
In plane PQ
line through P J_ CD,
construct
a
"" 148, 149.
5. Pi? is the perpendicular
required.
II.
Proof
Argument
1.
the
Through
in
P
Pig. 2)
2.
^P
3.
.-.
EF
plane PQ.
A. plane PQ.
4.
.-.
EF
_L
5.
But
6.
.-.
J_
III.
in
(P
plane
PR
in
Fig. 1,
MN,
draw
II4J3.
EF
PR
of
foot
PP
J_
PR
1.
; t\e.PR
_L
#*\
CD.
plane ifi\T.
The discussion will be
q.e.d.
givenin "
639.
as
PP.
314
SOLID
GEOMETRY
*
XIII.
Proposition
Two
640.
plane
are
straight
parallel.
lines
Theorem
perpendicular
to
the
same
*
rN
M
str. lines A B
Given
To
The
_L
CD
proof is
left
Hint.
Suppose
J5, as BE,
is IICD.
as
that
is not
AB
Use
If
||CD, but
of
the
the
that
line
plane
XIV.
is
with
given line is parallel
plane
To
prove
The
Hint.
plane MN.
through
tersecti
parallel to a plane, the inplane passing through
any
to the
given line.
and
plane AD, through AB,
secting
inter-
in line CD.
MN
II CD.
AB
proof is
line
\B
IIplane MN,
line AB
other
Theorem
Ax
Given
some
" 638.
straight
a
for the student.
exercise
an
Proposition
641.
plane MN.
II CD.
AB
prove
and
left
Suppose
that
as
an
AB
exercise
is not
for the student.
IICD.
Show
that
AB
will then
meet
642.
// a plane intersects
must,
if sufficiently
I.
Cor.
lines, it
and
AB
plane
does
in
MN
it,then
let it intersect
hut
CD,
IICD,
EF
if MN
Now
EF.
intersect
not
-parallel
plane through
a
and
CD
two
also.
the other
Pass
of
one
sect
inter-
extended,
Hint.
315
VI
BOOK
641.
"
is IIto
Apply
" 178.
643.
Cor.
If
n.
lines
tersecting
in-
two
each
are
parallel to a given plane, the
parallel to the given
of these
plane
lines
is
plane.
If
Hint.
determined
is not
plane
by
plane BS,
IIto
intersect it in
EF.
What
of EF
to AB
644.
line,as
is the
relation
and
CD
lem.
Proba
given
to construct
(a) A
(b) A
R
:
line
parallel to a given plane.
plane parallel to a given plane.
(a) Let A be a point outside
plane intersectingplane MN
Hint,
any
have
?
III.
Through
645.
CD,
it will
some
Cor.
point
MN,
and
AB
Cor.
their
of
plane
MN.
in line CD.
Through A
Complete the
struct
construction.
con-
If two angles, not in the same
plane,
parallel respectively,their planes are
IV.
sides
parallel.
Ex.
on
1173.
Hold
the blackboard
Ex.
1174.
a
pointer parallelto
parallelLo
Find
point and parallelto
the
a
locus
the
pointer
of all
given plane.
the
?
blackboard.
why
Is its shadow
?
straightlines passing through
a
given
316
GEOMETRY
SOLID
XV.
Proposition
If
646.
straight lines
two
other.
the
one,
a
is
plane
MN
containingCD.
IIAB.
Argument
1.
2.
plane MN is IIA
Suppose MN is not IIAB
3.
Then
4.
This
5.
.-.
plane MN
is
Cor.
I.
Through
648.
Cor.
a
space.
II.
or
it is not.
; then
1.
" 161, a.
2.
" 629.
3.
" 642.
CD.
4.
By hyp.
q.e.d.
5.
" 161, 6.
plane MN
AB.
must
also intersect
MN
Problem.
CD.
contains
IIAB.
plane parallel to
Hint.
B
for
impossible,
plane MN
647.
a
Reasons
Either
will intersect
taining
plane conparaliel to
#
CD, and
and
plane MN
prove
only
,
IIlines AB
Given
To
4
parallel,
are
lines, and
of the
one
Theorem
Through a given
another
given line-
E, any
point in CD,
Problem.
plane parallel
construct
Through
to any
two
a
line
to construct
line HK
IIAB.
given point
given straight
a
to
struct
con-
lines in
317
VI
BOOK
XVI.
Proposition
Theorem
If two straight lines are intersected by three parallel
of these lines
planes, the corresponding segments
proportionalytf
y
649.
are
"
/
IIplanes MN,
Given
A, E,
and
B
line
PQ,
and
in C, H, D,
CD
AE
intersectingline
RS
in
AB
respectively.
CH
"
To
prove
=
"
"
"
EB
HD
Reasons
Argument
.
1.
Draw
2.
Let
AD
the
intersect PQ
3.
Let
the
4.
.\
and
AE
AF
EB
FD
CH
EB
HD
in
RS
in
by
and
FH
CH
AF
HD
FD
and
AD
MN
in
"" 612, 616.
3.
"" 612, 616.
4.
" 633.
5.
" 410.
6.
"54,1.
AC.
Q.E.D.
650.
2.
BD.
IIAC.
FH
,
AE
and
EF
determined
plane
IIBD
EF
6.
in
intersect PQ
DC
1. " 54, 15.
intersectingplane PQ in F.
plane determined
by ^LBand^D
If two straight lines are intersected by three
divided
parallel planes, the lines are
proportionally.
Ex.
cut
are
Cor.
If any
1175.
by
two
or
more
In
the
number
of
lines
passing through
a
common
point
parallelplanes, their corresponding segments
are
proportional.
Ex.
CD.
=
1176.
28.
Find
AF
figure for Prop. XVI,
and
HD.
AE
=
6, EB
=
8, AD
=
21,
318
GEOMETRY
SOLID
Proposition
A
651.
line
straight
is
planes
XVII.
Theorem
perpendicular
to the
perpendicular
to
of
one
other
two
allel
par-
also.
N
r
7
R
B
plane MN
Given
To
line
prove
JL
AB
line
and
IIplane RS
In
2.
Then
3.
But
4.
.*.
AB
5.
.-.
line
AB
Cor.
I.
652.
one
plane MN, through
determined
plane
RS.
Only
C, draw
by ^Cy,and
CD
any
intersect
line
let the
CD, and
plane
in
RS
EF.
IIEF.
CD
J_
AB
J_ plane
plane MN.
Argument
1.
AB
EF.
_L CD, any
JL
line in
plane
C.
MN.
Through
plane parallel
passing through
plane MN
to
q.e.d.
given point there
given plane. (Hint.
exists
a
a.
Apply
only
""638,
6446, 651, 625.)
653.
"" 6446
Through
parallel to
654.
a
a
Cor.
plane, they
655.
of the
656.
of the
Def.
and
652
may
given point there
exists
in
one
and
one
statement
only
:
plane
one
given plane.
II.
If
are
parallel to
The
planes
two
projection
perpendicularfrom
Def.
be combined
The
the
projection
of
projections
all
each
of
each
are
a
other.
point upon
point to
of
a
pointsof
parallelto
line
the
upon
(Hint.
a
See
third
" 180.)
plane is the
plane.
a
a
plane is
the line upon
the
foot
the locus
plane.
319
VI
BOOK
A
XVIII.
Proposition
The
657.
projection
perpendicular
not
str. line AB
Given
To
the
prove
plane of a straight
plane is a straight line.
a
upon
to the
not
Theorem
plane MN.
_L
of
projection
AB
upon
MN
a
str. line.
Reasons
Argument
1.
2.
3.
1.
" 639.
2.
"" 612, 616.
3.
" 179.
4.
" 636.
plane
5.
" 655.
plane
6.
" 656.
7.
Args. 2
Through C, any pointin AB, draw CD _L
plane MN.
Let the plane determined
by AB and CD
in the str. line EF.
intersect plane MN
From
H, any point in AB, draw HK, in
plane AF,
4.
Then
5.
.-.
K
IICD.
is the
plane MN.
projectionof
is the
of
projection
HK
_L
line
R
upon
MN.
6.
.-.
EF
AB
upon
MN.
7.
.'.
the
projectionof
is
str. line.
a
AB
upon
plane Jf^T
and
q.e.d.
Compare the length of the projection of a line upon
plane with the length of the line itself :
(a) If the line is parallelto the plane.
(6) If the line is neither parallelnor perpendicular to the plane,
*^c)If the line is perpendicularto the plane.
Ex.
v
1177.
6.
320
SOLID
GEOMETRY
Proposition
all
Of
658.
I. Those
XIX.
oblique lines
Theorem
drawn
from
point
a
to
a
plane
having
equal projections are equal.
II. Those
having unequal projectionsare unequal,
the one
having the greater projection is the longer.
line
Given
J_
PO
Oblique
I.
:
and
plane MN
and
:
lines
PA
and
PB
with
projectionOA
lines
PA
and
PC
with
projectionOC"
=
tion
projec-
OB.
Oblique
II.
tion
projec-
OA.
To
The
659.
lines
I.
II.
the
:
prove
I.
proofis
Cor.
drawn
PB
left
I.
as
an
PC
exercise
"
PA.
for the student.
(Converse of Prop. XIX).
from
a
point
to
plane
a
Cor.
II.
the
Cor.
plane is
662.
of the
:
The
greater projection.
locus
of a point in
circumference
from all points in the
straight line perpendicular to the
and
passing through its center.
661.
Of all oblique
Equal oblique lines have equal projections.
Unequal oblique lines have unequal projections,and
longer line has
660.
II.
PA;
=
the
Def.
III.
The
shortest
line
perpendicular from
The
distance
perpendicularfrom
from
the
a
of
a
plane of
from
that
point to
point to
equidistant
space
the
circle
the
is
a
circle
point to a given
point to the plane.
a
a
plane is the
plane.
length
322
SOLID
GEOMETRY
DIHEDRAL
666.
A
Defs.
planes
that
dihedral
diverge
is the
angle
from
called
angle are
planes,its edge.
ANGLES
line.
a
The
its faces, and
figure formed
by two
hedral
planes forming a di-
the intersection
of these
dihedral
angle may be designated by reading in
order the two planes forming the angle; thus,an angle formed
by planes AB and CD is angle AB-CD,
If
and is usuallywritten angle A-BC-D.
there is no other dihedral
angle having
the same
edge,the line forming the edge
is a sufficient designation,as dihedral
angle BC.
667.
A
668.
Def.
lying in
Points, lines, or
the
said to be
plane are
same
planes
coplanar.
A
669.
of
dihedral
a
to the
first to have
670.
The
Def.
formed
by
figurewe
axis to the
an
of
plane angle
about
revolved
positionof
in the
5Cas
has
in the
Thus
been
about
of them
one
two.
by imagining that
portionsof planes, were
finite
as
*
magnitude
be obtained
angle may
faces, considered
coplanar and that
revolved
of the
clear notion
and
AB
positionof
in each
then
face
EF, in face AB, is _L
FH, in face CD, is J_
then
Z
is the
Ex.
1188.
All
Ex.
1189.
Is the
BC?
Ex.
1190.
?
Prove.
plane
of
State your
309
vertices
true
of
a
dihedral
angle
EFH
result in the
angle
Thus
BC
at F,
Z A-BC-D.
angle are
equal.
(" 667) perpendicular to
form
if
of
a
the
theorem.
quadrilateralis a quadrilateralin
quadrilateralare not all in the same
if the
the
point.
same
of the dihedral
plane angles of
Is Ex.
i.e. if the
space,
plane
Prove.
F, and
at
plane Z
EFH
edge
BC
to have
face of the dihedral
at
the
CD
CD.
straightlines,one
angle, perpendicular to its edge
two
mon
com-
face
angle is the
dihedral
a
line
a
first
at
imagine
may
face
its two
BOOK
671.
; thus
them
and
edge
common
a
dihedral
Two
Def.
Z B-CB-E
if
adjacent
are
face
which
they
have
lies between
and
D*
dihe-
adj.
are
angles
common
a
Z A-BC-D
323
VI
dral A.
672.
these
angle, and
be
,
said
are
perpendicular
to each
if
to
as
dral
adjacent diheangles equal, each of
anglesis a right
planes
to
so
two
dihedral
the
plane
one
another
meets
make
If
Def.
other.
plane
plane
meets
HP
LM
that
so
dihedral
and
Thus
A
H-KL-M
M-LK-N
are
equal, each
Z
is
rt. dihedral
Z,
and
with
plane geometry, frame
terms
in
acute
dihedral
and
planes HP
By comparison
1191.
Ex.
a
the
are
other.
_L to each
definitions of the
definitions of the
exact
dihedral
corresponding
followingterms :
dihedral
angle ;
angle ; reflex
dihedral
angles ; complementary
obtuse
angle ;
LM
oblique dihedral angle ; vertical
angles ; supplementary dihedral angles ; bisector of a
angle ; alternate interior dihedral angles; corresponding dihedral
Illustrate
Ex.
as
many
If
1192.
adjacent dihedral
Hint.
Ex.
See
1193.
rightdihedral
Hint.
Ex.
of these
See
1194.
as
plane
angles is two
one
'Hint.
See
with
an
open
another
angles.
book.
plane, the
rightdihedral angles.
meets
dihedral
sum
of the
two
proof of " 65.
If the
of two
sum
angles,their
proof
If two
adjacent dihedral
exterior
faces
are
angles is equal to
coplanar.
two
of " 76.
planes intersect,the
equal.
"
you
can
hedral
di-
proof of "
77.
vertical
dihedral
angles
are
324
SOLID
If
673.
angles
A
To
XX.
dihedral
angles
Theorem
are
equal, their
and
b'o' whose
equal dihedral A BO and
M'n'o',respectively.
Z
prove
MNO
BO
edge
edge
shall fall upon
Likewise
5.
.'.A
674.
_L
MNO
Cor.
"
54, 14.
The
plane
2. " 670.
pointN.
JV'O' are
collinear.
M'N'O'.
Z
I.
lines in
collinear.
are
and
NO
=
at
7iC
J/V
and
4.
m'n',two
and
are
.'. MN
that
1.
B'O'.
MN
3.
B'o',so
its
equal,
point N of
point N' of
upon
Z
AB,
a
Z
dihedral
Then
2.
is
Reasons
Superposedihedral
BO
are
M'N'O'.
Z
=
plane A
Argument
1.
-plane
equal.
two
MNO
Proposition
two
are
Given
GEOMETRY
q.e.o.
plane angle of
a
right
3.
" 62.
4.
""
5.
"18.
670, 62.
dihedral
angle
right angle.
675.
Cor.
to
third
plane
If
II.
a
intersecting planes are
third plane, their intersections
two
intersect
each
other.
each
with
pendicular
per-
the
planes
_L plane
Given
and
CD
AB
MN
intersectingeach
and
other in line
let
and
AE
; also
DB
be the
FC
intersections of
and
AB
325
VI
BOOK
planes
plane
with
CD
MN.
To
and
that
prove
AE
each
intersect
FC
other.
Reasons
Argument
1. Either
each
2.
3.
4.
IIFC
AE
intersect
FC
161, a.
1.
other.
2. "
Then through 77, any
Suppose AE IIFC.
point in DB, pass a plane J/fi7" _L 7A7,
in 2,.
in if and AE
FC
intersecting
is _L ;(# also.
"
Then
plane HKL
is the plane Z of dihedral Z
7/"X
"
.*. Z
hedral
.FC, and Z ifLTJ is the plane Z of diZ
5.
and
AE
or
636.
670.
"4#.
dihedral
But
627.
A
and
FC
AE
dral
rt. dihe-
are
" 672.
5.
A.
.'.A
HKL
and
.'.A
HKL
contains
this is
But
AE
Ex.
1195.
points in
Ex.
and
are
rt. A.
"674.
two
rt. A.
Arg. 6.
JTL7J
"206.
impossible.
intersect each
FC
Find
the
locus
Find
the
locus
Are
the
other,
of all
" 161, b.
q.e.d.
points equidistant from
two
given
three
given
space.
1196.
of all
points equidistant from
points in space.
Ex.
1197.
complements
Ex.
can
1198.
they
be
?
Prove
If
your
two
parallelto
other, prove
ea'qh
supplements
of
equal
dihedral
angles equal?
answer.
planes
each
are
other?
their intersections
Explain.
with
the
If
third
third
plane,
they
parallelto
plane parallel.
perpendicular
each
to
a
are
326
SOLID
GEOMETRY
Proposition
XXI.
Theorem
(Converseof Prop. XX)
If the
676.
-plane
equal, the dihedral
M'N'O'
To
angles
A BC
dihedral
two
Given
angles of
dihedral
two
are
A
and
equal.
are
b'c' whose
and
angles
plane
MNO
equal.
are
dihedral
prove
A
BC
dihedral
=
Z B'o'.
Reasons
Argument
1. Place
dihedral
Ab'c'
A
that
so
superposed
BC
plane A
its
upon
dihedral
upon
shall
MNO
1.
"
2.
" 670.
3.
" 622.
4.
" 638.
5.
" 612.
6.
"18.
54, 14.
be
equal, plane
Am'n'O'.
2.
3.
Then
.-.
B'c'
and
BC
NO
at N.
BC
and
B'c'
are
are
both
both
_L
_L
and
MN
plane MNO
at N.
4.
.-.
BC
5.
.-.
planes
MN
CD
NO,
6.
.-.
^'c'
and
and
are
determined
jl'JJ',
and
AB
and
collinear.
are
coplanar;
J5C, are
C'D',determined
coplanar.
dihedral
A
BC
=
dihedral
also
by
Z
by
planes
and
BC
#'c'.
Q.E.D.
BOOK
327
VI
If the plane angle of a
angle
right angle, the dihedral
angle.
677.
1199.
Prove
Ex.
1200.
If two
alternate
dihedral
the
plane and
1200, and
Ex.
cut
State
it
prove
If
1202.
the
by
of
converse
XI
Ex.
a
/y
ly
J jS V
f/C /
/D
zf
parallel planes
1/
are
y1
transversal
plane, the ing
corresponddihedral angles are equal. (Hint.
See proof
by
/
_-
Ex.
47
JmL
/
Prove
633).
j"-
/
lie in
CD
JW
f
the indirect method.
two
plane, the
XL
plane
*/
1201.
Ex.
transversal
jj
T-X.
M-Z
a
plane
plane Z of dihedral
is the
DCB
Z
II("
are
a
equal.
the
and
AB
by
cut
are
intersect
BO
Then
CD.
same
that
Let
and
by AB
is
right dihedral
a
plane Z of
the
V-WX-Y.
Z
determined
angles are
be
is
angle
by applying " 676.
parallelplanes
Z ABC
Let
in
1194
interior dihedral
Hint.
Z
Ex.
Ex.
MN
dihedral
Cor.
1203.
State the
1204.
If two
of Ex.
converse
1202, and
of
" 190.)
prove
it by the indirect
method.
Ex.
Ex.
are
1205.
either
dihedral
Two
equal
faces
angles whose
supplementary dihedral
or
by
a
transversal
plane, the
side of the transversal
angleson the same
angles. (Hint. See proof of " 192.)
rightdihedral
plane is two
cut
are
dihedral
interior
of the two
sum
parallelplanes
are
each
parallel,
to
each,
angles. (Hint. See proof of
" 198.)
Ex.
1206.
A
angle has the
dihedral
plane angle. (Hint. Proof
Ex.
1207.
dihedral
Two
numerical
same
similar to that of "
angles have
the
measure
as
its
358.)
same
ratio
as
their
plane
angles.
Ex.
points
1208.
not
Find
a
lying in the
Ex.
1209.
If
Ex.
1210.
If
point
plane.
in
a
plane equidistant from
three
given
straightline intersects one of two parallelplanes, it
prolonged, intersect the other also. (Hint. Use the
must, if sufficiently
and
indirect method
apply "" 663 and 664.)
a
a
plane intersects
one
extended, intersect the other
sufficiently
and apply "
lfte'^iod
652.
)
of two
also.
parallelplanes,it must, if
(Hint.
Use
the
indirect
328
SOLID
GEOMETRY
Proposition
XXII.
Theorem
If a straight line is perpendicular to a plane,
plane containing this line is perpendicular to the
678.
every
given plane.
El
M
plane MN and plane PQ
and intersecting
plane MN in CD.
To prove
plane PQ _L plane MN.
str. line AB
Given
AB
_L
containingline
Argument
1.
AB
_L
2.
Through B,
3.
Then
Z
4.
But
5.
.-.
"619.
Z
in
plane MN,
CD.
2.
"63.
is the
ABE
of dihedral
3.
"670.
draw
plane Z
BE
_L
Q-CD-M.
Z ABE
Z,
and
a
Z
Q-CD-M
plane
a
is
the
a
a
a
line is drawn
point
rt. dihedral
plane ifJV.
line in the
any
from
"619.
the foot of
If from
rightangles to
line drawn
rt. Z.
is
plane PQJL
1211.
to
"677.
q.e.d.
pendicular
perat
plane, the
of
tion
intersec-
point in the perpendicular
is perpendicular %o the line of
formed
so
the
1.
CD.
dihedral
Ex.
Eeasons
to
any
plane.
Hint.
Ex.
JEK=
Make
1212.
KE
In
10, find AK.
=
the
EH.
Prove
figureof
Ex.
AK
=
AH,
1211, if AB
and
=
apply "
20, BE
142.
4V11,
and
330
SOLID
GEOMETRY
Proposition
681.
If each
to
I. Their
line
Their
line
II.
third
of intersection
of intersection
other
in line
3.
is
perpendicular
I. That
:
prove
_L
RS
plane
MN
and
AB
intersects
AB
the
each
intersecting
J_
plane MN.
Argument
intersect
plane MN in
lines PD
and RE, respectively.
Then
and RE intersect in a point as C.
PD
intersects
.*. AB
through C; i.e. AB
passes
Let
and
planes PQ
RS
q.e.d.
II.
Argument
1. Either
_L
AB
Suppose
other
line
3.
Then
4.
.-.
CF
5.
.*.
planesPQ
CF
or
1.
but that
some
through C, the point common
planes,is _L plane MN, as line
lies in
is
it is not.
plane MN,
plane PQ,
is the intersection
which
.-. AB
plane MN
is not _L
AB
the three
6.
to
plane MN.
plane MN.
2.
plane.
AB.
I.
2.
and
planes PQ
II.
1.
the third
intersects
plane.
Given
To
Theorem
of two intersectingplanes is perpendicular
plane ;
third
a
XXIV.
and
RS
of
also in
to
OF.
plane RS.
planes PQ
intersect in two
and
str.
RS.
lines,
impossible.
plane MN.
q.e.d.
6. "
161,
b.
BOOK
XXV.
Proposition
any
to construct
plane,
plane.
a
To
plane perpendicular
a
to the
to
given
plane MN.
through AB, a plane _L plane MN.
_L
not
AB
construct,
The
perpendicular
not
^,5
line
Given
Problem
straight line,
Through
682.
331
VI
construction,
proof,and
discussion
left
are
as
an
cise
exer-
for the student.
Hint.
Apply " 678.
683.
to
the
discussion,see
Through
Cor.
plane, there
a
For
" 683.
straight line, not perpendicular
only one plane perpendicular to
a
exists
given plane.
Hint.
MN.
would
What
follows
Ex.
one
a
and
1216.
Ex.
683
about
may
AB
plane through
AB
"
plane
?
be combined
in
one
statement
as
straightline,not perpendicularto a plane, there
only one planeperpendicularto the givenplane.
Apply
intersect at the
book
know
exist another
:
Through
exists
you
"" 682 and
684.
should
there
Suppose
corner
the
of
truth
a
room
placed perpendicularto
1217.
plaaes,it is
the
of Prop. XXIV
;
(6)
to the
a
(a)
planes
to the
planes
that
formed
by
of two
intersecting
an
open
top of the desk.
plane is perpendicular to
perpendicular to their intersection.
If
:
each
332
SOLID
T^
GEOMETRY
Proposition
685.
To construct
in
straight lines
a
XXVI.
Problem
perpendicular
common
To
AB
and
construct
X
str. lines in space.
two
CD, any
a
two
space.
K
Given
to any
line _L both
to AB
and
to CD.
I. Construction
1.
Through
Through
CD
construct
plane MN
IIAB.
AB
construct
plane AF
_L
in EF, and
CD
in
2.
MN
3.
Through
4.
HK
H
plane
intersecting
MN
" 682.
H.
HK, in
construct
is J_ to both
" 647.
^5
and
CD
plane AF,
and
is the
_L
" 148.
EF.
required.
line
Reasons
1.
By
cons.
"641.
By
cons.
"193.
"679.
" 619.
7.
in
III.
The
686.
Cor.
the
same
discussion
Between
plane)
will be
two
there
given in "
straight
exists
Args. 4
and
6.
686.
lines
only
one
in
space
common
{not
pendicular.
per-
ZW?
draw
Y
Through
CD?
to
relation of XB
the
686
"" 685 and
687.
follows
two
there exists
one
A
1218.
a
lower
CD.
AB?
to
to
What
EF.
is
statement
one
as
extends
plane, the
unequal
The
1220.
planes
two
either the
the
;
other
two.
Ex.
(6)
drawn
from
on
Find
of any
an
has
high.
floor and
walls
the
length of
along
the
tion
intersec-
given plane
a
given
a
If two
equal.
?
by
two
parallellines
line which
is
parallelto each
determined
in
are
the greater inclination
their projectionson
parallel,
a
Prove.
and
a
of the
plane
two
planes is perpendicular to the other
of the planes is perpendicular to the
illustration of this exercise
If two
1223.
are
planes
parallel,no
are
is
two
:
third
perpendicular to
the
in the classroom.
line in the
one
meet
can
other.
Find
1224.
all
Find
1225.
from
feet
Find
points equidistantfrom
parallelplanes
two
equidistant from three points : (a) if the points lie
(b) if the points lie in one of the planes.
Ex.
10
given point to
a
of the three lines of intersection
each
line in the
Ex.
are
of three
If each
intersection
(a)
plane
be
and
wide,
line, or parallellines.
same
1222.
Ex.
lines
If two
1221.
can
lines to the
point not in their plane, intersect
given parallels.
Ex.
feet
drawn
are
drawn, which
thus
are
15
ceiling.
of these
inclinations
lines
line that
equal lines
If two
1219.
Ex
plane)
same
perpendicular.
diagonallyopposite corner.
the floor,then
diagonallyacross
walls to the
of two
the
(not in
to the
corner
line that
space
long,
feet
shortest
from
any
in
be combined
may
is 20
room
Find
Ex.
MN
and
JL to AB
of XY
relation
"
XB
Through X draw
?
Complete the proof.
straightlines in
and only one common
the length of the
the
second
a
:
Between
Ex.
plane MN?
to plane
is the
What
||AB.
ZW
to
figure of " 685,
in
Suppose JI,
Hint.
333
VI
BOOK
two
all
points equidistantfrom
parallel planes, and
at
a
given
in
and
plane
neither
;
given points,equidistant
two
distance
d
from
a
third
a
given
plane.
Ex.
line,the
Ex.
1226.
If each
intersection
1227.
of two
of the
intersectingplanes
planes
is parallelto the
Construct,through
shall be parallelto two
a
point
intersecting
planes.
is
parallelto
line.
in space,
a
straightline
that
334
SOLID
GEOMETRY
XXVII.
Proposition
plane BE
Given
and
AC
To
the
and
AC
Js from
PK
Z
P, any
the
pointin
2.
/%.
Reasons
planes^
_L
4.
.-.
BC"
5.
.-.
Apch
planes AC
C and
plane
CD
and
dihedral
J_
are
and
CD;
"
2.
" 678.
3.
"681,11.
4.
" 619.
5.
" 670.
6.
By hyp.
7.
" 673.
8.
Byiden.
9.
" 209.
10.
" 110.
planeMN.
UK,
CH, CP, and
the dihedral
6. But
_L
plane MN
BC
.\
612, 616.
1.
in C.
Then
i.e.
by planes
plane BE,
PK.
Through PH and PK
pass plane MN
intersectingplane AC in CH, plane
in Off,plane BE
in PC, and edge
CD
BC
angle'
formed
Argument
1.
dihedral
a
respectively.
CD,
PH=
prove
and
bisects
faces of
dihedral
bisectingthe
; also PH
CD
to faces
plane that
equidistant from
is
angle
the
in
Every point
688.
Theorem
C^.
jBTCP
are
A E-BC-A
Z
E-BC-A
the
plane z"
and
=
of
D-CB-E.
dihedral
Z D-CB-E.
Zpch=Zkcp.
7.
.-.
8.
Also
9.
.*.
rt. A
10.
.'.
PH=
PC
=
PC#=
PK.
PC.
rt. A
KCP.
Q.E.D.
BOOK
689.
Cor.
faces of
angle.
a
335
VI
Every point equidistant from the two
dihedral
angle lies in the plane bisecting the
I.
plane bisecting a dihedral
angle is
the locus
of all points in space equidistant from the
faces of the angle.
690.
Cor.
n.
691.
Cor.
m.
The
given dihedral
bisector of
a
angle.
Prove
1228.
Ex.
the
construct
To
Problem.
that
a
dihedral
angle
can
be bisected
by only
one
plane.
See
Hint.
proof of "
Find
1229.
Ex.
Find
1230.
Ex.
lines.
intersecting
of all
locus
many
points equidistantfrom
planes does
this locus consist
the locus of all points in space
Of how
planes does
many
the locus of all
Find
1231.
Ex.
the
Of how
planes.
599.
space
?
equidistantfrom
this locus
points in
secting
inter-
two
two
consist ?
equidistantfrom
two
parallellines.
Find
1232.
Ex.
locus of all
points in space equidistantfrom two
equidistantfrom all points in the circumference
the
planes and
intersecting
of
a
circle.
Ex.
Find
1233.
the locus of all
Find
1234.
intersecting
planes,equidistantfrom
from
fixed points.
two
Ex.
If from
1235.
supplementary
is
planes,M
such
Ex.
MN.
Given
1236.
Ex.
that
any
and
N.
PX+XQ
Find
two
Find
is
Given
1237.
a
point
Ex.
a
two
X
minimum.
"fyNT. See
point
faces of the
perpendicularto the
points in
equidistantfrom
planes and
intersecting
Ex.
locus of all
the
175.
to the
space
fixed
two
equidistantfrom
points.
two
points in space equidistantfrom two
two
parallelplanes,and equidistant
within
a
dihedral
angle
angle, the
#, one
the
by
of the dihedral
plane angle
points,P and
a
point Xin
angle lines
formed
in each
intersection
are
the
drawn
diculars
perpen-
angle.
of two
of
intersecting
planes M and N
minimum.
points,P
in plane
and
MN
Q,
such
on
one
that
side of
PX
+
a
XQ
given plane
shall be
a
336
SOLID
GEOMETRY
POLYHEDRAL
692.
A
Def.
ANGLES
polyhedral
is the
angle
figuregenerated by
a
moving straightline segment that continuallyintersects the
boundary of a fixed polygon and one extremityof which is a
fixed point not in the plane of the given polygon. A polyhedral
called a solid angle.
angle is sometimes
693.
generatrix,
the
694.
called
the
polygon ABODE;
the
the
of
vertex
is called
a
fixed
polyhedral
,
V.
of the
through
the
are
generatrixin
The
Defs.
elements
gon
v
polygon
I
element
an
line is called the
moving
; the fixed
VA
as
directrix, as
point is
angle,as
is
The
Defs.
of the
vertices
VA,
as
/\/
poly-
Bf"
tt(
the
(
\
etc. ;
VB,
portions of the planes determined
by the
edges of the polyhedral angle, and limited by
faces, as AVB, BVC, etc.; the angles formed
by
the
face
formed
angles,
A
as
AVB,
by the faces are
angle,as dihedral
695.
A
face
sometimes
together are
696.
The
Def.
etc.; the
BVC,
called the
A
angles
and
may
them
the
the
are
edges
dihedral
angles
the
hedral
poly-
the
dihedral
of
a
angles taken
polyhedral
be designated
by
a
angle.
letter at
each
If there is
edge, as V-ABCDE.
other polyhedral angle having the same
vertex, the letter
the vertex
is a sufficient designation,as V.
the vertex
and
one
on
are
etc.
called the parts
polyhedralangle
\ Af
angles of
dihedral
VA, VB,
J
the
polyhedralangle;
the
edges,
position 4/
any
angle is
no
at
polyhedralangle
whose
directrix is a convex
polygon,i.e. a polygon no side of
In
which, if prolonged,will enter the polygon; as V-ABCDE.
this text only convex
polyhedralangles will be considered.
697.
Def.
698.
Defs.
directrix
is
A
convex
A
polyhedral
trihedral
angle
is
a
a
polyhedralangle
whose
dral
triangle(tri-gon)
angle, a polyhe; a tetrahedral
angle whose directrix is a quadrilateral(tetra-gon)
; etc.
a
338
SOLID
GEOMETRY
XXIX.
Proposition
trihedral
Two
704.
angles of
equal if the three face
are
equal respectivelyto the three face
other, and the equal parts are arranged in
one
angles of
the
the
order.
same
trihedral
Given
Theorem
A
angles
are
Z A'V'b'
V'-A'B'C',Z AVB
c'v'a',and the equalface angles
and
V-ABC
=
,
Z
VC=Z
B
B'v'c',Z
arranged in
To
the
CVA
order.
same
trihedral
prove
Z
=
Z
Outline
of
Z
trihedral
V-ABC=
v'-A'b'c'.
Proof
Since, by hyp., any two face A of V-ABC, as A AVB
and BVC, are equal,respectively,
to the two
corresponding face
A of v'-A'b'c',it remains
dral
only to prove the included dihe1.
A
2.
Let
face A
and
meet
Since
and
AVB
and
ED
A
and
AVB
in
VC
and
D
Similarly,
lay off
Prove
rt. A
DVE=vt.
Prove
rt. A
EVF=
Prove
A
.
A
be
BVC
EF, in
705.)
"
obliqueA\ then from any
planesAVB and BVC, respectively,
VB.
and
VA
4.
_L
(See also
equal. " 702, II.
in VB, draw
point E
3.
v'b'
and
VB
DEF
FVD
=
=
A
obliqueA, ED
respectively.Draw
BVC
are
F,
V*E'
=
A
rt. A
A
f'v'd'
D'E'F'
:
VE
and
draw
A
and
EF
will
FD.
D'e'f'.
D'V'E' ; then
VD"
V'd',ED=E'd'.
then
E'v'f'-,
VF=
EF=E,f'.
v'f',
;
then
then
Z
FD
DEF
=
=
f'd'.
Z
D'E'F'
A
But
9.
10.
.-.
dihedral
11.
.-.
trihedral
705.
Z
rt.
are
dihedral
706.
"
plane A
of dihedral
A
VB
704.
A
are
Z and
707.
if their
reverse
XXIX
homologous
plane
dihedral
face
A,
is true
Z
as
if
of the
Z
a
dral
trihe-
of the included
and
VB
only one
other
are
V'B',
face Z
rt.
A,
propositionin Bk. I that corresponds to
step in the proof of that proposition? Did
main
dihedral
Z
VB
of
" 704
polyhedral angles
Two
Def.
Prop.
is the
of
A
State the
correspond to proving
that
face Z
face A
If two
equal.
q.e.d.
that all the dihedral
show
A,
homologous
that
its
are
F'-^'.B'C'.
Z
oblique.
the
was
two
rt.
are
that all
to prove
Questions.
What
face A
hence
v'b'.
Z
trihedral
F-XBC==
that
hence
of the first trihedral
dihedral
=
that the third
show
A,
It remains
if all face
VB
Z
and
A
Z, and
equal.
Z
If all the
Note.
dihedral
rt.
are
or
the
are
r'B',respectively.
and
are
d'e'f'
and
DEF
339
VI
BOOK
corresponding parts are
are
=
dihedral
said
to
Z
V'B' ?
be
metrical
sym-
equal but arranged in
order.
By making symmetrical polyhedralangles and comparing
them, the student can easily satisfyhimself that in general
they cannot be made to coincide.
708.
if the
Def.
Two
edges of each
polyhedral angles are said to be vertical
the prolongations
of the edges of the
are
other.
It will be
seen
that two
like two symmetrical,polyvertical,
hedral
their
corresponding parts equal but
angles have
order.
arranged in reverse
Two
"
."
Equal Polyhedral
Angles
Two
Vertical
hedral
PolyAngles
Two
Symmetrical PolyAngles
hedral
340
SOLID
GEOMETRY
Proposition
I.
the
If
of
angles
Theorem
are
symmetrical
the two
and
The
left
proofs are
Hint.
Let
arranged in
what
V
and
reverse
V'
as
be
order.
exercises
the
two
Construct
will be the relation of V"
1240.
Can
two
trihedral
trihedral
to
V
polyhedral angles
Ex.
Are
two
Ex.
1241.
Are
be
tively
respec-
:
order.
reverse
with
Z
V"
to
equal
parts
symmetrical
to
but
V.
V?
symmetrical
and
equal ?
If two
polyhedral angles
if symmetrical, are
they
?
two
trirectangulartrihedral angles necessarilyequal?
birectangulartrihedral angles equal ?
1242.
A
? of V
equal ? symmetrical and vertical ?
vertical,are they necessarilysymmetrical ?
necessarilyvertical
equal
for the student.
vertical and
are
and
the
included
dihedral
face angles and
one
are
equal respectivelyto two face angles
the included
dihedral
angle of the other ;
If the three face angles of one are
to the three face angles of the other
provided the equal parts are arranged in
Ex.
gles
an-
two
III.
Then
:
adjacent dihedral
one
are
equal respectivelyto a face angle
adjacent dihedral
angles of the other;
If
angle of
and
face angle
a
two
II.
trihedral
Two
709.
XXX.
If two
trihedral
angles have
Prove
three face
your
answers.
angles of
equal
respectivelyto three face angles of the other, the dihedral
angles of the
firstare equal respectivelyto the dihedral angles of the second.
one
341
VI
BOOK
/
XXXI.
Proposition
The
710.
any
greater than
is
angle
of
sum,
Theorem
trihedral
face angles of a
the third face angletwo
V
c
trihedral
Given
Z
is
Z
in
V-ABC
Z
prove
BVC
Z
+
Z
"
CVA
Outline
1. Iii
face
through
2.
On
But
5.
In
6.
But
7.
.'.
A
+
FG
A
GE
and
GVE
Z
Z
FVG
FVG
+Z
J^FG
711.
used
here
1243.
Ex.
=
difference
ADVF;
"
FD
+
"FD,
draw
and
then
DE;
GE
.:
that
in E
and
and
A
BVC,
VB
in
and
F.
GE.
D#.
"
GVE
=
FD.
=
Z
"
"Z#F.Z)
"
J"FD.
give the
ZZ)^;
^FS.
Q.E.D.
in
theorem
the
theorem
Z
+
be
statement
Any
of the other
face
two.
as
angle
to the
of
a
number
trihedral
I that
Bk.
proved by
proof.
If,in trihedral angle V-ABC,
a
DVB
FG
^(?
Z
prove
CTJ.
State
If so,
80",make
1244.
VD
=
Z
+
Can
?
Z
making
intersectingva
=
(?FE
Question.
XXXI.
Prop.
VG
Proof
Z.DVF.
=
i.e. ZBVC
Ex.
greatest face
AVB.
of
VD
line
any
lay off
VC
4.
draw
AVB
draw
D
3. Prove
CVA
the
AVB.
To
one
which
a
corresponds
similar
method
to
to
the
60",and angle
angle BVC
of degrees in angle AVB.
=
angle
is greater
than
the
342
SOLID
GEOMETRY
XXXII.
Proposition
The
712.
sum
is
angle
Theorem
of all the face angles of any convex,
less than
four right angles-
hedral
poly-
V
polyhedralZ
Given
To
the
prove
Hint.
Let
etc.
From
O,
any
How
many
A
have
all the
Z
ABV
Z
+
of the
sum
with
F,
FBO
base
A
with
Z
the
713.
sum
Question.
to
nearly
most
A
Prop.
1245.
Ex.
Can
triangles?
Ex.
1246.
Can
1247.
Can
a
is the
vertices at V?
at
O ?
Which
is the
a
+
Z
at
07? O ?
vertices
is
at
which
Then
F,
or
the
greater, the
sum
greater,
greater, the
is the
A
base
of the
of A
about
face A
of the
sum
of
sum
0 ?
?
a
proposition in plane geometry
If so, state
it.
If not, state the
sponding
corre-
one
that
it.
polyhedral angle
four ?
.
What
there
corresponds to
.
0 ?
about
Is
polyhedral Z in A, B, C,
draw
OA, OB, OC, etc.
.,
of the
ABC
4 rt. A.
Z
V
at
which
XXXII
at
V?
with
Then
of the
faces.
edges
polygon
ABO
of A
vertices at O ?
or
in
their vertices
or
A
n
intersect the
point
of all the
A
with
of the face A
sum
plane
a
V
five ?
for its faces
have
three
lateral
equi-
six ?
polyhedral angle
have
for its faces three squares
?
four ?
Ex.
pentagons
Ex.
can
possiblybe
and
have
for its faces
three
regular
Show
that
formed
Can
with
the greatest number
regularpolygons
trihedral
of
as
polyhedral angles that
faces
is five.
angle have for its faces a regulardecagon
angle,
equilateraltriangles? a regular decagon, an equilateraltri? two
a
regular octagons and a square ?
square
1249.
two
polyhedral angle
? four ?
1248.
Ex.
and
a
a
VII
BOOK
POLYHEDRONS
714.
finite
A
solid
surface
and
Def.
closed
716.
A
Def.
surface
bounding
717.
bounding
planes
finite
planes only.
intersections
called
are
of
the
dron
is
vertices
closed
edges,
A
is called
a
one
of
many
How
Ex.
1251.
What
Ex.
edges
1253.
1254.
faces, and
Exs.
result
of six
least
tetrahedron
of faces
number
faces, a
decahedron
do-
; etc.
icosahedron
a
hexahedron;
a
of twelve
one
faces,an
faces,
y
?
that
a
hexahedron
a
?
polyhedron
can
?
How
many
edges
How
many
vertices
has
a
tetrahedron
?
hexahedron
a
?
an
?
octahedron
Ex.
vertices
1252.
octahedron
Ex.
?
"
four
has
"i
"|pi"11
one
diagonals
is the
whose
/wl"ffl*^
/$lM
jSfflll
IIlllllfllfc
jM^WtSSm^
^^M^^Mmm^W
^|["W
AB.
octahedron;
twenty
B
two
of
tetrahedron;
eight faces, an
1250.
?
face, as
same
polyhedron
Ex.
have
polyhe-
a
it.
by
//\
the
of
the
of
figure
;
the
edges,
portions
diagonal
in the
;
in
edges
straight line joining any
Defs.
of
one
a
not
719.
faces
A
Def.
of
of the
the
by
bounded
space
solid
faces, of the polyhedron.
718.
an
portion of
of
composed
a
composed
space
is
bounding planes bounded
the
figurein
a
separates
space.
a
the
and
vertices;
the
remaining
figure is
closed
if it
closed
is
intersections
the
the
from
be
to
polyhedron
The
Defs.
is said
surface
of space
portion
715.
a
A
Def.
Fthe
1252
is known
has
a
tetrahedron?
a
hexahedron?
?
If
E
represents
number
and
as
1253,
Euler's
the
number
of vertices
show
in
that
each
in
theorem.
343
each
of
of the
case
edges,
the
F
polyhedrons
E+2
=
of
number
mentioned
V+F.
This
344
SOLID
Ex.
where
'
8 is the
Does
hexahedron
720.
whose
?
an
faces
angles and
?
regular
a
Fis
the
4
dodecahedron
?
is
polyhedron
How
Questions.
4
(V"2)
=
(V"2)
formula, 8=
octahedron
8
right angles, hold
equilateraltrianglescan
many
right angles,
of vertices.
number
for
polyhedron all of
all of whose
dral
polyhe-
a
equal regularpolygons,and
equal.
are
angles are
721.
tetrahedron
a
face
the
A
Def.
in
that
of the
sum
1256.
Ex.
a
Show
1255.
GEOMETRY
meet
to
form
of regular
what
is the greatest number
polyhedral angle (" 712) ? Then
is the
polyhedrons possiblehaving equilateraltrianglesas faces ? What
of regularpolyhedrons possiblehaving squares
faces ?
as
greatest number
a
having regular pentagons
faces
the
regular polygons of
From
722.
the
faces ?
more
of kinds
number
maximum
drawn
as
than
of
Can
a
regular polyhedron have
five sides ?
that
not
more
exist.
than
He
convince
should
polyhedrons
five are
possibleby actuallymaking them
indicated
below
Tetrahedron
What, then, is
?
regular polyhedrons possible?
questionsin " 721, the
the conclusion
why
student
has
doubtless
five kinds
of
himself
that
from
regular
cardboard
:
Hexahedron
as
Octahedron
Dodecahedron
Icosahedron
these
as
346
SOLID
728.
a
plane
and
A
Def.
which
which
right
GEOMETRY
of
section
prism is
a
is
perpendicularto
the lateral edges or
cuts
a
lateral
a
the
section formed
by
prism
edge of the
edges prolonged.
rA
\
\
Right Prism
729.
A
Def.
right prism is
perpendicularto
730.
Oblique
Regular Prism
whose
lateral
edges
are
whose
bases
are
the bases.
A
Def.
prism
a
Prism
regular prism is
rightprism
a
regularpolygons.
731.
An
Def.
oblique to
are
732.
as
733.
any
the
The
(c) Any
(d) TJie
(e) The
(f) The
(g) TJie
is
plane of
should prove
two
lateral
a
edges
prism is the perpendicularfrom
base to the
one
of the
some
"prism
edges of a
edges of a prism
lateral
edge of a rightprism
lateral
faces of a prism
lateral faces
bases
of
sections
a
plane of
are
of a prism
prism ;
:
parallel.
are
is
equal
to the altitude.
parallelograms.
of a rightprism
prism
of each
a
base.
equal.
are
are
the other
propertiesof
the correctness
lateral
all the lateral
base
of
altitude
followingare
(h) Every
lateral
whose
ing
prism is triangular, quadrangular, etc.,accordetc.
are
triangles,
quadrilaterals,
The
Def.
(a) Any
(b) The
prism
A
its bases
the student
a
the bases.
Defs.
pointin
734.
oblique prism is
rectangles.
are
equal polygons.
made
by
two
parallelplanes cutting
edges are equal polygons.
section
equal to
of
the base.
a
2^'ism made
by
a
plane parallelto
the
BOOK
Proposition
735.
Two
prisms
placed,to
Theorem
and
equal respectively,
are
one
three
I.
equal if three faces including
are
angle of
trihedral
347
VII
faces including
a
trihedral
a
larly
simi-
angle of
the other
A
prisms
Given
A'G',face
To
B
AD
and
AI
face
A'l',
AJ
face
=
A'j',face
Argument
FAE, and
.
3.
\
trihedral
Place
prism
trihedral
4.
Faces
A
BAE
b'a'e'.
A
AI
upon prism A'l1 so that
shall be superposed
A
=
trihedral
equal, trihedral
and
J, AG,
AD
" no.
equal,respectively,
are
Z
Z
its
upon
face
Reasons
b'a'f',f'a'e',and
to A
2.
=
prism A*f.
prism AI=
1. A BAF,
AG
face A'd'.
=
prove
B
A
are
Z
Z
A1.
"704.
"
54, 14.
A'.
spectively,
equal,re-
and A'D'.
A'j',
A'G1,
fall upon J\ F1,and
By hyp.
to faces
5.
.*.
J, F, and
will
G
5. "18.
G',respectively.
and
c'h'
G.
CH
7.
.*.
CiJand
8.
.-.
H
9.
Likewise
both
are
C'h'
will fall
are
" 734, a.
||BG.
collinear.
" 179.
upon#'.
2" will
fall upon
" 603, b.
l\
ilar
By steps simto 6-8.
prism
10.^.-.
AI"
prism A'l'.
Q.E.D.
10.
" 18.
348
SOLID
736.
A
Def.
made
but which
737.
the base and
by
trihedral
are
738.
Two
I.
Cor.
Ex.
Ex.
section of the
a
truncated
Two
and
Two
1257.
equal, each
cluded
in-
prisms
three
H.
bases
equal
prism
faces including a
are
angle of one
spectively
equal reto three
faces including a
the
angle of the other, and
similarly placed.
trihedral
faces
a
plane obliqueto the base,
all the edges of the prism.
cuts
Cor.
portion of
a
equal if
are
prism is the
truncated
between
prism
GEOMETKY
to
right prisms
are
equal if tlieyhave
altitudes.
equal
triangularprisms
equal if their
are
lateral faces
are
each.
faces are :
angles
Classify the polyhedrons whose
(a) four triand
three
two
triangles
parallelograms; (c) two quadri(6)
laterals
and
four
two
and
four
parallelograms; (d)
quadrilaterals
1258.
;
rectangles; (e)
Ex.
edges are
prism. (Hint.
lateral
Ex.
1260.
edge
is
a
two
sum
lateral
the
Draw
a
rt.
four
a
rectangles.
of the
plane angles of the
edges of a triangularprism ;
section of the prism.)
Every section of
parallelogram.
a
made
prism
a
made
prism
by
by
dihedral
a
angles
quadrangular
a
plane parallelto
a
a
plane parallelto
a
parallelogram.
The
1262.
through
the
and
Every section of
1261.
lateral face is
Ex.
squares
Find
1259.
whose
Ex.
two
section
of
a
parallelopipedmade
diagonallyoppositeedges
is
a
by a plane passing
parallelogram.
"3
iiiiiiiiiiiiiiF
Oblique Parallelopiped Right Parallelopiped
Rectangular
Cube
Parallelopiped
739.
Def.
A
parallelograms.
parallelopiped
is
a
prism
whose
bases
are
740.
lateral
edges
741.
right parallelopiped is
A
Def.
A
whose
parallelopiped is
rectangular
bases
whose
parallelopiped
a
the bases.
to
perpendicular
are
Def.
349
VII
BOOK
right parallelopiped
a
rectangles.
are
a regularhexahedron)is
(i.e.
whose
edges are all equal.
parallelopiped
742.
Def.
743.
The
A
cube
followingare
should
student
; the
of the
some
of
properties
a
opiped
parallel-
of each
the correctness
prove
lar
rectangu-
a
:
(a) All the faces of a parallelopipedare parallelograms.
(b) All the faces of a rectangularparallelopipedare rectangles.
(c) All the faces of a cube are squares.
(d) Any two oppositefaces of a parallelopipedare equal and
parallel.
(e) Any
as
oppositefaces of
two
a
parallelopiped
may
be taken
the bases.
faces are : (a) six parallelograms
Classifythe polyhedrons whose
six
six
two
parallelograms
rectangles; (c)
squares
; (d)
; (")
and four rectangles; (e) two rectanglesand four parallelograms; (/) two
squares and four rectangles.
Ex.
1263.
Ex.
1264.
Find
the
Ex.
1265.
Find
the
Ex.
1266.
Find
the
edges are
(5,8,
1267.
Ex.
of the cube
and
The
of all the
sum
diagonal of
12 ; whose
of
edge
; find
=l:x:y
edges
a
x
cube
and
a
angles
", and
a,
c.
the
diagonal of
face
a
Ex.
1269.
The
diagonals of
Ex.
1270.
The
diagonals of
a
parallelopipedbisect
Ex.
1271.
The
diagonals of
a
parallelopipedmeet
Ex.
with
Ex.
point is sometimes
1272.
1273.
The
in the
sum
1274.
Is the
statement
is 20 V3
a
rectangularparallelopipedare
of the
the
surface, is bisected
parallelopipedis equal to
Ex.
whose
diagonal
cube
called the center
of the
diagonal
the
a
Any straightline through
its extremities
:
y.
Find
This
e.
rectangular parallelopipedwhose
1268.
edge
parallelopiped.
a
is 8 ; 12 ;
edge
whose
are
:
of
Ex.
the
of
cube
a
diagonal of
face
squares
the
sum
in Ex.
of
at the
true
a
equal.
other.
point.
a
parallelopiped,
center.
diagonals of
of the squares
1273
in
d.
parallelopiped.
center
of the four
each
;
a
of the twelve
for any
gular
rectan-
edges.
parallelopiped?
350
SOLID
GEOMETRY
PYRAMIDS
744.
A pyramidal
Def.
surface
is a surface
generated by
intersects
straightline that continually
and that passes through
in the
a fixed point not
planeof
745.
the broken
694, give
of
the
and
tions
defini-
generatrix,
and
directrix, vertex,
of
element
surface.
out
a
pyramidal
Point
in the
746.
these
figure.
A
Def.
idal
pyram-
surface
of
two
consists
parts lying on
opposite
sides
vertex, called
and
lower
747.
Def.
of
the
the
per
up-
nappes.
A
pyramid
is
polyhedron whose
a
of the
portion of a pyramidal
vertex
to a plane cuttingall
its elements,and the section
formed
by this plane.
748.
to
give
the
surface
sists
boundary con-
extending from
its
referring
student
may
the definitions of base,
lateral
of
By
Defs.
727,
"
line
ferring
re-
"" 693
to
fixed broken
a
ing
mov-
line.
By
Defs.
a
faces, and
lateral
edges
of
pyramid. The vertex
the pyramidal surface is
as
V.
a
Point
In this text
these
out
in the
only pyramids
will be considered.
called
the
vertex
of the
pyramid,
figure.
whose
bases
are
convex
polygons
BOOK
749.
750.
Can
these
may
a
bases
751.
the
used
be
terms
of
altitude
a
triangularpyramid
interchangeably?
have
pyramid
a
in the
How
?
a
rahedron
tetferent
dif-
many
?
is the
plane of the base, as
figureon precedingpage.
to the
vertex
below, and
faces has
many
triangularpyramid
The
Def.
cording
etc.,ac-
etc.
a quadrilateral,
triangle,
a
How
Questions.
?
from
is
its base
as
is triangular, quadrangular,
pyramid
A
Defs.
351
VII
perpendicular
in the figure
VO
V
Truncated
Regular
Pyramid
752.
Pyramid
A
Def.
erected
753.
to the
base
Def.
A
included
by
a
A
Def.
included
made
by
755.
a
The
the student
is
Regular Pyramid
pyramid
a
whose
lies in the
vertex
of
base
is
a
perpendicular
at its center.
the
plane cuttingall
754.
whose
truncated
between
Frustum
Triangular Pyramid
regular pyramid
regular polygon, and
of
Frustum
between
and
base
the
a
section
portionof a pyramid
of the pyramid made
edges.
of
frustum
is the
pyramid
the
a
pyramid
base
and
a
is the
portionof
a
mid
pyra-
section of the pyramid
plane parallelto the.base.
followingare
should
prove
some
of the
the correctness
of
properties
of each
a
pyramid;
:
(a) TJie lateral edges of a regularpyramid are equal.
(b) TJie lateral edgesof a frustum of a regularpyramid are equal.
(c) The lateral faces of a regularpyramid are equal isosceles
triangles.
(d)
The
lateral
faces of
.isoscelestrapezoids.
equal.
a
frustum of a regularpyramid
are
352
SOLID
GEOMETRY
II.
Proposition
756.
If a -pyramid
I.
The
edges
II.
The
section
the lateral
To
edges
i.
:
prove
a
are
polygon
similar
and
plane
ABODE
V"
divided
in F, G, H, I, and
"
VD_VC_
VF
VG~
J
to the base
:
proportionally.
to the
base.
||base
MN
AD
cutting
the altitude in P.
and
_
FGHIJ~
VH~
VQt
VP
ABODE.
Argument
I.
2.
is
plane parallel
a
~
II.
1.
by
altitude
and
pyramid
Given
is cut
Theorem
Through V pass plane RS
Then
plane RS IIplane MN.
VA
3.
VB
VB
_
VO
VO
VH'
VH
_
VF~
VO' VG~
_VC
VA_VB
4.
IIplane KL
_
VH~
VF~~ VG~
757.
base
proof of
The
II.
Cor
is to the
vertex
Hint.
Any
I.
base
is to the
II is left
section
as
the
square
as
of
VB2
a
square
of the
Prove
AB1
exercise
an
VO1
for the student.
parallel to the
distance
from the
pyramid
of
its
altitude
of
the
pyramid.
354
Prove
1281.
Ex.
other
the
as
The
1283.
Ex.
feet and
square
450
altitude of the
of the
a
of
sides
frustum
10
are
the
pyramid.
?
regular hexagonal pyramid
must
area
frustum
a
of
of
the
of
a
is 12 V3
altitude
plane
be
passed
?
pyramid
a
frustum
the
given figureis
288
are
is 3 feet.
a
frustum.
lateral
regular pyramid are equi18 inches, respectively;
and
inches
Find
is 8 inches.
each
to
are
of the
vertex
vertex
a
frustum
a
of which
pyramid
of
bases
feet ; the
bases
pyramid
a
the
a
pyramid of which
triangleswhose
the
of
base
section whose
square
The
1284.
from
far -from the
of the
areas
altitude of the
the
Ex.
to form
base
the
How
of
fulfillthis condition
1280
side of
Each
parallelto the
distances
in Ex.
altitude is 15.
is 6 ; the
Eind
their
of
squares
1282.
Ex.
parallelsections
that
results obtained
the
Do
GEOMETRY
SOLID
tude
alti-
the
given figure is
a
frustum.
Ex.
1285.
pyramid
is
and
VO
be
altitude of the
OE
or
of any
the base.
than
figure,let VE
the
greater, VE
the
lateral faces
of the
sum
greater
In the
Hint.
VAD
The
the
altitude of face
pyramid.
Which
is
?
MENSURATION
OF
THE
PRISM
AND
PYRAMID
Areas
760.
of
turn
761.
The
Def.
a
pyramid
In
notation
6, c
dimensions
=
is the
a
of the
sum
of the
will be used
of
of
area
mensuration
the
lowing
a,
lateral
prism, a pyramid, or
of its lateral faces.
areas
prism
=
area
of base in
lower
b
=
area
of
a
base
upper
or
base
frustum.
of
H
h
L
=
lateral
of
or
edge
in
general.
a
or
a
element,
altitude of
a
solid.
=
altitude of
a
surface.
=
slant
height.
a
right
of
the
VltV2
=
lateral
T
=
total
V
=
volume
...
=
tion
sec-
lower
frustum.
base
frustum.
a
S
tetrahedron
=
of
of
perimeter of upper
of
edge,
the fol-
pyramid.
a
perimeter
frustum.
E
of
generalor of
of
base
pyramid
vertex
parallelopiped.
B
and
:
lar
rectangu-
a
frus-
a
area.
area.
in
volumes
into
general.
of smaller
which
solid is divided.
a
ids
sol-
larger
III.
Proposition
of
the
lateral
The
762.
lateral area, and
To
S
prove
the
P
P
=
a
Theorem
a
prism
right
section
with
prism AK
Given
of
area
perimeter of
355
VII
BOOK
MQ
rt.
a
is
and
a
lateral
rt. section
section
2.
.'.MN"AI;
3.
.-.
.-. area
area
etc.
altitude
the altitude
4.
Reasons
NQ".CJ\
is the
MN
the
E.
"
_L Al, CJ, etc.
MQ
edge,S
MQ.
Argument
1. Rt.
product
edge.
lateral
a
section,E
perimeter of
to the
equal
of
O
of O
CK;
etc.
AI=
MN-
CJ=
NQ
of O
AJ
=
JW-
of CJ
CK
=
NQ
"
is
NQ
AJ\
E-,
"
1.
" 728.
2.
"619.
3.
" 228.
4.
"481.
5.
"54,2.
6.
"309.
E-,
etc.
5.
CJAJ+CJCK-]
6.
.*.
8
763.
the
=c
P
"
+ JTQ +")*"
=s(JCir
#.
The
Cor.
product of
Hint.
764.
Thus,
of any
765.
Def.
the
if P
The
Def.
one
lateral
=
of a right prism is equal to
perimeter of its base and its altitude.
perimeter of
slant
of its
The
is the altitude
area
height
and
base
of
a
H
=
altitude,S
regular pyramid
=
P
is the
"
H.
tude
alti-
triangularfaces.
slant
height
of any
one
of
a
frustum
of its
of
a
regular pyramid
faces.
trapezoidal
356
SOLID
GEOMETRY
Proposition
The
766.
lateral
IV.
of a regular pyramid is equal to
of the perimeter of its base and
area
half the product
its slant height.
one
C
D
with
the perimeter of its
regularpyramid O-ACD
denoted
by P, its slant height by L, and its lateral area by S.
Given
base
To
"
S
prove
767.
a
equal
the
of
The
side
=
Find
of
a
square
height.
L.
\(P
+
the
lateral
p)
area
and
regular pyramid
each
base
is 24
inches,and
altitude is 16 inches.
Ex.
1287.
pyramid
slant
area
of whose
whose
8
of
area
perimeters of
its slant
and
1286.
total
L.
lateral
the
of
Prove
Ex.
.
""
regular pyramid
half the product
a
one
sum
Hint.
the
of
to
its bases
"-P
=
Cor.
frustum
is
Theorem
are
The
15
sides of the
centimeters
height is 30 centimeters.
the lateral
area
of the frustum.
bases
and
Find
24
of
a
frustum
of
a
regular octagonal
centimeters, respectively,and
the
number
of square
decimeters
the
in
Ex.
Find
1288.
quadrilateralwith
lateral
the
sides
of
area
and
5, 7, 9,
357
VII
BOOK
prism whose
a
inches,and
13
rightsection is
lateral
whose
a
is
edge
15 inches.
Ex.
inches
and
Ex.
whose
base
base
is 10 V3
Find
triangularbase is
Find
1292.
the
altitude is 20 inches and
one
1293.
Find
the total
Ex.
1294.
Find
the
Ex.
; if its total
1295.
of
of
edge of
meters
prism is 45 deci-
a
lateral
a
edge makes
side
one
is 195 square
area
base
cube
a
13 inches.
with
of whose
inches.
regular hexagonal prism
a
whose
is 10 inches.
diagonal is 8 VS.
whose
if its total
cube
a
of
altitude is 16
area.
lateral
area
area
11, and
regular prism,
a
side of whose
Ex.
centimeters
of
whose
total
; and
lateral
the
and
5 inches
sides 8,
right section
a
altitude
rightprism whose
a
decimeters
Find
the
of
trianglewith
a
angle of 60".
1291.
Ex.
is
perimeter of
altitude
an
Ex.
lateral area
the
The
1290.
; its
the
Find
1289.-
is 294
area
square
is T.
area
Find
the total
Find
the
of
area
regular tetrahedron
a
whose
edge
is 6 inches.
Ex.
1296.
whose
Ex.
slant
pyramid,
Ex.
a
Find
6, 8, and
are
Ex.
whose
base
square
1300.
Ex.
whose
bases
is 4
The
inches,
bases
Ex.
bases
bases
lateral
area.
Ex.
an
are
4
of
regular
a
hedron
tetra-
regularhexagonal
altitude
whose
b, and
a,
is 10 inches.
c.
side
rightparallelopiped,one
a
of
altitude is 6 inches.
theater
is
Find
they
a
rectangularparallelopipedwhose
whose
a
of
area
supported by
the
20 feet
are
cost,
at
high
and
In
are
In
the
angle of 60".
and
12
a
a
lateral
a
frustum
a
of
2
columns
four
cents
the
a
square
apothems
regular triangular pyramid,
and
inches,respectively,
height and
1303.
inches,and
and
and
a
of
area
if
frustum
a
8 and
1302.
of the
8
In
the slant
Ex.
of
area
of
10 inches.
1301.
of the
Find
are
total
total
are
regular hexagons.
are
foot, of painting the columns
the
edges
total
balcony
of
area
12 ; whose
the
and
area
is 6 inches
base
the total
Find
1299.
and
area
8 inches.
the lateral
side of whose
1298.
edges
height is
Find
1297.
lateral
of
a
the
regulartriangularpyramid,
lateral
sides
regular hexangular pyramid, the sides
and
8, respectively,
the
the
altitude is 10 inches.
edge.
altitude is 12 inches
Find
the
area.
;
a
altitude is 16.
the
Find
side of the
lateral face makes
with
the
base
the
is
base
358
GEOMETRY
SOLID
Volumes
768.
The
Note.
student
corresponding discussion
the
A
769.
solid
it contains
a
should
of the
compare
rectangle,""
be measured
may
solid unit.
The
carefully"" 769-776
with
466-473.
by finding how
solid unit most
many
times
frequentlychosen
is of unit
length. If the unit length is
an
inch,the solid unit is a cube whose edge is an inch. Such
If the unit length is a foot,the
inch.
a unit is called a cubic
solid unit is a cube whose
edge is a foot,and the unit is called
is
a
a
cube
cubic
whose
foot.
Fig.
770.
which
volume
771.
edge
Def.
Rectangular ParallelopipedAD
1.
The
is called
the
result
of
the
measure-number,
60
=
is
measurement
or
77.
numerical
a
number,
measure,
or
of the solid.
Thus, if
the unit cube
U
is contained
in the
rectangular
D
K
Fio. 2.
Rectangular ParallelopipedAD
=
24
IT-
BOOK
359
VII
AD
(Fig.1) 60 times, then
parallelopiped
the measure-number
or
rectangularparallelopiped
AD, in terms of U, is 60.
If the given unit cube is not contained
in the given rectangular
of times without
a
mainder
reparallelopipedan integralnumber
(Fig.2),then by taking a cube that is an aliquot part
of U, as one
to the
eighth of U, and applying it as a measure
will be obtained
rectangularparallelopiped
(Fig.3),a number
volume
Fig.
of
3.
Rectangular ParallelopipedAD
=
""*
U+
=
39|
U+.
which, divided by 8,* will give another (and usually closer)
of the given rectangularparallelopiped.
approximate volume
By proceeding in this way (Fig.4),closer and closer approximations
to the true
Fig.
"
4.
may
be obtained.
Rectangular ParallelopipedAD
*
I
volume
It takes
eight of
the small
cubes
to make
=
*%%"
the unit
U+
=
cube
TJ+.
41T93
itself.
360
GEOMETRY
SOLID
If the
772.
the unit
edge'of
the
and
is
which
be found
given rectangularparallelopiped
of the
edges
cube
commensurable,
are
of
may
will be
which
and
cube
a
U,
aliquotpart
an
integralnumber
rectangularparallelopiped
in the
an
tained
con-
of
times.
If the
773.
edge of the unit cube
and
the
and
closer
cube
no
given rectangularparallelopiped
of the
edges
approximationsto
incommensurable, then
are
be obtained,but
may
U will be also an
aliquot
the volume
aliquotpart of
(by definition
rectangularparallelopiped
is
which
part of the
closer
an
mensurable
of incom-
magnitudes).
is,however, a
closelyby
There
and
more
smaller
subdivisions
774.
The
is incommensurable
approach
parallelopiped
solid,is used
to
a
volumes
The
Def.
ratio
Two
Def.
of any
volumes
measure-numbers, or
776.
is the
the
of
volume
the volume
mean
these
as
rectangular
of the unit
of
cube
solid,or
simply
solid with
respect
a
of the
limit
unit.
chosen
775.
to
by using
cube,
cube
subdivisions
the
as
unit
unit
chosen
approach zero as a limit.
For brevitythe expression the
the
obtained
more
limit.
approximate
.
the
of
approached
which
rectangularparallelopiped
a
the
with
successive
a
as
of
volume
is
approximations
approach zero
Def.
which
the
subdivisions
smaller
and
definite limit which
solids
two
(basedon
equivalent
are
ratio of their
is the
solids
the
same
unit).
if their
volumes
are
equal.
777.
Historical
polyhedrons
which
as
is
3400
b.c.
of
where
thought
an
in
a
a
c
the
document
copy
In
of
a
this
determination
as
ancient
of
as
the
manuscript dating back
manuscript
of the formula
barn
Ahmes
volumes
the
Rhind
cannot
tested.
papyrus,
calculates
"
be
of
possiblyas far
of the formula, V"
a
by means
be
linear
of
the
to
dimensions
are
supposed
exact
shape of these barns is unknown,
Egyptian
unfortunately
accuracy
to be
(See " 474.)
a, b, and
The
Note.
is found
b
"
the
(c + \ c),
barn.
so
sents
con-
that
But
the
362
GEOMETRY
SOLID
(b) Suppose
but
is not
that
u
that
some
left
The
proof is
II.
If AC, AF, and
as
a
of AC, AF, and
measure
aliquotpart
exercise
an
AG
each
are
of
u
is such
Let
be
m
incommensurable
with
of
times
as
many
will be
There
3.
Now
hence
4.
.-.
the volume
Now
take
matter
AQ
with
the
of
=
a
how
AC, AF,
"627.
3.
"337.
4.
"
5.
"335.
m.
"
MK
AC,
_L AF, and
_L AG.
each
are
measure
mensurable
com-
m,
and
AM
AN
"
"
of
measure
small
a
AG, the
a
No
u.
of
measure
it is appliedas
and
778, I.
AQ.
u
is
measure
remainders,
MC, NF, and
QG, will be smaller than
the
taken.
measure
2.
MC,
as
orectangularparallel
smaller
taken,when
to
and
"339.
u, the linear unit.
with
piped AK
5.
draw
AN,
AM,
plane
plane NK
plane QK
draw
1.
possible.
as
less than
draw
M
Through
through.N
through Q
as
m
AG, respectively,
remainders,
QG, each
NF, and
2.
Apply
u.
to AC, AF, and
measure
a
u.
Reasons
measure
a
measure.
for the student.
Argument
1.
a
AG,
spectively
re-
BOOK
363
VII
Reasons
Argument
6.
.-.
the
difference
between
AM
and
AC,
the
difference
between
AN
and
AF,
and
the
AG,
may
and
between
difference
each
be
made
less than
remain
7.
8.
.-.
.-.
approaches AC
as
approaches AF
approaches AG as a
AM
as
9.
AN-
"
as
a
limit,AN
limit, and AQ
a
AF
difference
the
between
-
AG
and
5.
tangular
rec-
be
may
less than
remain
and
Arg.
gular
rectan-
AD
parallelopiped
made
to become
any
previously assigned volume,
small.
however
the
"593.
limit.
limit.
a
Again,
.*.
8.
small.
parallelopipedAK
10.
"349.
previously
approaches AC-
AQ
7.
become
to
any
AM
Arg. 5.
and
AQ
however
assigned segment,
6.
volume
of
rectangularparallelopiped
AK
approaches the volume of
rectangular parallelopipedAD as a
10.
"349.
11.
Arg.
12.
"355.
limit.
11.
But
the volume
to AM-
12.
.-.
III.
AN
of
of AD=AC-AF"AG.
If AC
and
AF
with
The
of
proofs of
779.
Cor
its
edge.
q.e.d.
is commensurable
If AC
with
IV.
always equal
I.
H*N". Compare
4.
AQ.
"
the volume
is
AK
with
u
but
AF
and
AG
are
commens
in-
u.
commensurable
are
with
u
but
AG
is incommens
u.
III and
The
with
IV
volume
" 478.
are
left
of
a
as
exercises
cube
is
for the student.
equal
to the
cube
364
GEOMETRY
SOLID
780.
as
Note.
the
By
with
Compare
(Hint.
781.
two
other
each
to
are
rectangular
parallelopipeds
mensions
the products of their three di-
Any
Cor.,II.
product
of
surface
a
of the
of the measure-numbers
product
" 479. )
and
surface
a
and
the
line is meant
the
line.
of a rectangular parallelopiped is equal to the product of its base and its altitude.
783.
Cor. IV.
two
rectangular
parallelopipeds
Any
782.
Cor.
their
other
each
to
are
The
III.
volume
with
Compare
(Hint.
altitudes.
their
products of
the
as
bases
and
" 479.)
rectangular parallelopipedshaving
equivalent bases are to each other as their altitudes; (b)
two
rectangular parallelopipeds having equal altitudes
784.
Cor.
other
to each
are
785.
dimensions
in
dimensions,
and
two
having
786.
here
VII
other
rectangular
common
as
peds
parallelopito each
are
their
other
dimensions.
two
IV
Book
is it in
the
Ex.
1305.
The
volume
of the square
side
Ex.
of the
1306.
10 and
Ex.
cube
whose
large;
The
The
.
1308.
Ex.
base.
is
of
that
corresponds to
volume
1309.
{.e.
The
of two
edge
is twice
of
that
edge of
containing twice
The
of
area
volume
diagonal is 5\/3 ; d.
whose
side of the
the
dimensions
total
cube
third the altitude of the
one
Find
a
rectangular parallelopipedis V; each
parallelopiped. Find
a
5, 12, 16, respectively.
1307
Ex.
base
of
volume
Find
8,
ing
hav-
? to
1304.
the
in
other
their
Ex.
side
" 480.)
and corollaries
rectangularparallelopiped? State the theorem
in Book
Will
IV that correspond to "" 778, 779,782, 783, and 784.
proofs given there, with the corresponding changes in terms, applywith "" 769-785.
?
Compare the entire discussion of "" 466-480
in Book
the
(b) two
each
to
are
common
What
Questions.
with
Compare
(Hint.
rectangular parallelopipeds
dimension
one
products of
the
as
bases.
their
as
(a) Two
VI.
Cor.
third
(a) Two
V.
a
a
a
the
cube
192
ratio of their
is 300 square
certain
of the
=
cubic
feet.
are
rectangular parallelopipeds
Find
cube
if V
base
cube
is
inches
V;
find
6,
volumes.
;
find its volume.
the
volume
of
a
given cube.
is
the volume
a
; find
of the
edge of
given cube.
the
a
cube
twice
as
BOOK
787.
Historical
discover
solution
a
double
There
are
the
is
volume
whose
of
that
finding of
the
of a cube, i.e.
edge of a cube
famous
that
to
given cube.
two
legends as
a
to
the
originof
the
problem.
one
is that
old
tragicpoet represented
as
wishing
to erect
a
an
King
Minos
tomb
for his
The
cus.
The
Glau-
son
fied
dissatis-
king being
with
dimensions
the
(100
feet each
his
way) proposed by
The
architect,exclaimed :
in-
"
closure
the cubical
The
the
small
is too
; double
tomb
for
fail not
it,but
legend asserts
other
feringfrom
who
of
plague
a
the
altar
was
constructed
The
pestilence became
was
It
was
its head.
this
Later
with
he
No
is noted
contributions
additions
"Let
of
form
long
eight years
would
have
cube.
a
A
new
that of the old
as
was
after
the
traveling and
in
a
pupil of
one.
the
pealed
ap-
Socrates.
to
his association
study
of
matics.
mathe-
with
acquainted
the
Archytas, who
these people that
u
a
380
About
Over
no
Depart,
It is stated
imporfcratas Pythagoras
on
an
improvements
are
He
of
valued
geometry
thinking
and
advance
enter
my
grip
of
was
door."
away
philosophy."
originaldiscoverer,and
in
in the
of Greek
rather
its method
mainly
good authority that
to the
the
gave
his school
to
actually turned
was
was
to his native
he returned
geometry
hast not
teacher, rather than
geometry
b.c
the entrance
geometry
for thou
with
as
a
in
of Socrates
death
became
he
that
ignorant
knew
right seeing and
processes."
for
school.
a
none
who
:
as
it was
mathematics.
to its matter.
in
time
doubt
applicant
the statement
Plato
this
established
inscription:
an
in
Delians
the
to
as
Pythagorean School, especiallywith
passion for
city,where
as
wealth, and
years
during
of the
his
some
twice
at Delos
before, whereupon the Delians
the Delian
as
problem.
known
and
considerable
spent
edge
than
worse
Athens,
in
he
members
its
having
oracle
the
Apollo replied that
in the
altar,which was
It is therefore
born
possessed
b.c.
him
Plato
suf-
typhoid fever, consulted
size of his
to Plato.
Plato
then
that
were
to double
399
in
stop the plague.
to
Plato
royal
a
form."
Athenians,
how
of the first to
one
(429-348b.c.)was
problem of antiquity,the duplication
Plato
Note.
365
VII
"means
his
than
ucation
of ed-
conception of imaginary
"Plato
was
geometry."
almost
as
366
GEOMETRY
SOLID
788.
whose
whose
Theorem
VI.
Proposition
oblique prism is equivalent to a right prism,
is a right section of the oblique prism, and
base
is equal to a lateral
altitude
edge of the oblique
An
prism.
o'
C
B
oblique prism
Given
ADf ; also rt.
prism EN'
base
with
EN
ad', and with EE', LL',etc., lateral edges
EN', equal to AA',bb', etc.,lateral edges of Ad\
To prove
rt. prism EN'.
obliqueprism AD*
a
rt. section
of
of
'o
Outline
1.
BK
In
truncated
prisms
to
equal,respectively,
2.
Prove
the
sides
Proof
of
and
AN
the A
of face
A'N', prove
the A
of
face
of face b'k'.
to
equal, respectively,
BK
the
sides of face b'k'.
.-.
4.
Similarlyface
5.
.-.
6.
7.
face
BK=
B'm', and
"ilf=face
truncated
prism AN
But truncated
prism A' If
.-. oblique prism AD'=ort.
789.
to
face b'k'.
3.
Prop.
Question.
VI
?
=
Is there
If not, formulate
a
BE"
face b'e'.
prism a'n' (" 737).
truncated
prism A'N.
truncated
=
prism EN'.
in
theorem
one
face
and
see
Book
if you
q.e.d.
IV
can
that
prove
corresponds
it true.
BOOK
Proposition
The
790.
the
volume
of
product
367
VII
VII.
Theorem
parallelopiped
of
any
its base and
is
equal
to
its altitude.
\2*M
m
its altitude
B, and
by
To
its volume
I with
parallelopiped
Given
V
prove
B
=
"
by
denoted
Prolongedge
2. On
Reasons
and
all
edges of
I IIAC.
prolongation of AC take DF
AC, and through D and F pass planes
77.
_L AF, forming rt. parallelopiped
=
I
Prolong edge FK
5. On
the
_L
and all edges of
prolongationof
FK, and
FN
,
FK
take
II IIFK.
MN
=
through M and N pass planes
forming rectangularparallelopiped
III.
6. Then
7.
8.
II
III.
o
.-.IoIII.
Again, BoB'
9. Also
=
b".
H, the altitude of 1,= the altitude
of ///.
10.
But
1-1*.-.
the volume
v=
B
"
H.
" 54, 16.
" 627.
" 788.
//.
=^
V, its base
H.
the
3. Then
4.
AC
by
H.
Argument
1.
S:K*j"n
of
111=
b"
.
H.
Q.E.D.
" 54, 16.
" 627.
368
SOLID
791.
Cor*
I.
Cor.
II.
GEOMETRY
Parallelopipeds having
equal altitudes are equivalent.
and
792.
other
the
as
793.
Cor.
bases
(b) two
each
J_ny two
products of
III.
(a) Two
are
to
794.
other
What
5 from
Ex.
1311.
the
The
If the
In
first the
the
an
Find
1313.
whose
Ex.
Ex.
by
a
Hint.
edge
solid,and
Ex.
4 feet 6
In
of
is
a
cube
cubic
inches,and
Ex.
1318.
the
same
inches,
volume
its
makes
with
a
the base
of the
in
four
rectangular parallelopipe
a
edges are
a,
volume
whose
eral
equal latangle of
60" ; and
volumes
equivalent to
cube
an
angle of
an
15 ; whose
b, and
is 512
In
a
Find
the
of
area
a
c.
cubic
section
are
of the box
is 3 inches
cement
inside
of
in the
certain
numerical
will be needed
feet of cement
made
make
a
inches,
thick ?
always find the volume
solid,then subtract.
of the whole
rectangular parallelopipedis 2430 cubic
Find
its edges.
ratio of 3, 5, and 6.
a
cube
value.
to
2 feet 6
are
of this kind,
volume
edges
a.
dimensions
if the
of the
The
1317.
cube
of
many
problem
a
adjacent
angle
diagonallyopposite edges.
two
How
the
a
diagonal
includinglid,if the inside
3 feet, and
have
of
and
equivalentbases
ratio of the
the
6, 10, and
are
the
inches.
plane through
1316.
edge
Find
edge
the
The
1315.
Ex.
box,
cubic
a
;
edges
Find
1314.
inches
the
IV
prisms
parallelogram two
a
of 45" ; in the third
of 90".
ume
vol-
proofs given there, with
and they include an
15, respectively,
parallelopipedis 10, find its volume.
lateral
angle
an
to
are
?
parallelopipedshave
Four
angle
parallelopipeds.
Ex.
the
parallelopipedis
a
8 and
are
30" ; in the second
fourth
of
base
altitude of the
1312.
edges.
the
altitudes
equal
Prop. VI by subtractingthe equal truncated
entire figure.
sides of which
Ex.
altitudes, and
Prove
Arg.
of 30".
their
as
Will
correspond to "" 790-793.
corresponding changes, apply here
of
altitudes.
IV
expression in Book
corresponds to
the
theorem
corollaries
in Book
and
Quote
that
1310.
their
bases.
parallelopiped?
a
and
each
to'
are
parallelopipeds having equivalent
each
their
as
Questions.
Ex.
bases
parallelopipeds having
other
of
parallelopipeds
their
bases
equivalent
the
Find
area
of
the
the volume
surface
and
of the
cube.
the
volume
370
SOLID
GEOMETRY
Proposition
797.
The
of
volume
the product
of
IX.
triangular prism
a
its base
Theorem
and
with
triangularprism A CD-X
V, its base by B, and its altitude by H.
798.
Can
?
figure CZ
A CD-X
Ex.
-
in
exercise
an
its volume
"
denoted
by
for the student.
propositionin
797 ?
Book
IV
is its volume
What
corresponds
of
name
CZ
part of
What
?
Prop.
to
is the
What
proof there given ?
the
apply
you
to
H.
What
Questions.
above
the
B
proof is left as
The
IX
V=
prove
equal
its altitude.
Given
To
is
is
(" 795) ?
oppositeedge
triangularprism
a
and
is
equal to
perpendicular from
the
half the
one
point in
any
the
to that face.
The
Hint.
of
lateral face
of any
product
volume
The
1319.
triangularprism
is
one
of
half
a
certain
parallelopiped
(" 795).
Ex.
sides 10
with
coal will the
Ex.
The
1320.
a
coal
Find
Ex.
face of
One
1321.
the volume
of the
1322.
During
will fall upon
gallonsto
Ex.
a
a
1323.
the number
\
inch
is 8 feet deep is
which
a
triangularprism
a
point
?
ton
oppositeedge
of
tons
many
inches
45 square
contains
in the
triangle
a
;
is 6 inches.
prism.
a
ten-acre
| inch, how many
counting 7" gallonsto a
field,
of water
barrels
rainfall of
cubic
foot and
31|
barrel ?
feet,2 feet 6
Find
bin
feet,15 feet,and 20 feet,respectively. How
bin hold considering35 cubic feet of coal to a
perpendicularto this face from
the
6
of
base
inches,and
of
thick,a
dimensions
inside
The
2
pounds
of
an
open
tank
the latter
feet,respectively,
of zinc
before
being
required to line the tank
cubic foot of zinc
weighing
6860
ounces.
lining are
the
with
height.
a
ing
coat-
X.
Proposition
of
of
volume
The
799.
Theorem
prism is equal
any
its altitude.
and
its base
371
vn
book
to the
product
^
AC
prism AM
Given
B, and
To
with
its altitude
V
prove
by
B
=
its volume
denoted
by
From
H.
.
Reasons
of the
vertex
any
lower
A, draw
2.
diagonalsAD, AF,
Through edge AI and these
pass planes AK, AM, etc.
3.
Prism
is thus
AM
by
H.
Argument
1.
V, its base
divided
base,as
1.
" 54, 15.
2.
" 612.
3.
" 732.
etc.
diagonals
into
lar
triangu-
prisms.
Denote
4.
the
volume
prism
by
ADF-K
Vj
5.
.-.
6.
.-.
^
+
ACD-J
v2 and
bxH ;
=
^2+
...
v2
=
Cor.
vl and
b2; etc. Then
b2H;
=
equal altitudes
801.
Cor.
n.
"797.
bx; of
etc.
" 54, 2.
(b1+ b2+-~)H.
are
" 309.
"
Q.E.D.
,
Prisms
I.
of triangular
by
Fs^.a
800.
base
and
equivalent
having
and
bases
equivalent.
Any
prisms
two
to
are
each
other
as
their altitudes.
products of their bases and
(a) Two
802. Cor. m.
prisms having equivalent bases
the
are
to
each
having equal
other
as
altitudes
altitudes:
their
are
to
each
other
(b) two
as
their
prisms
bases.
372
equal
Theorem
triangular pyramids having equivalent bases
Two
803.
XI.
Proposition
*
and
GEOMETRY
SOLID
altitudes
equivalent.
are
"
0
wN
/A
/
and
O'-a'c'd' with
triangularpyramids O-ACD
with altitudes each equal to QR, and
base A'c'd',
denoted
by V and V',respectively.
Given
ACD
=c=
volumes
To
V
prove
V'
=
.
Argument
V*,V
1.
V
2.
Suppose
=
"
V "
For
constant.
the two
are
3.
4.
5.
V'.
"
that
the
1.
v'
v=k,
"
a
plane,MN.
common
altitude
FGU, JEW,
with
into
QR
n
with
=
edges
QX.
prisms by II, III, etc.
On A'c'D',F'g'u',
etc.,as
construct
and
these
with
struct
bases,con-
etc.,as upper
altitudes
3.
"653.
4.
"759.
5.
"726.
6.
"726.
that their bases
equal parts, as QX, XY, etc., and
through the several pointsof division
pass planes IIplane MN.
Then
section
section
FGU
o
f'g'u',
section j'k' w', etc.
section JEW
o
On
2.
161, a.
" 54, 14.
convenience, place
same
prisms
G.
V
pyramids so
in the
Divide
Vf,or
V',so
Reasons
prisms
with
altitudes
=
IIDO
and
Denote
these
lower
bases,
edges
QX.
prisms by I',II'?etc.
IID'o'
Denote
base
with
BOOK
373
VII
Argument
Then
7.
II
prism
Reasons
prism II',prism
=c=
III
7.
"800.
of
8.
"54,3.
9.
12.
"54,12.
"54,9.
"54,5.
802, a.
13,
" 54, 10.
14.
Arg.
15.
ilar
By steps sim-
prism III',etc.
=c=
8. Now
the
denote
of the volumes
sum
prisms II, III, etc.,by 5; the sum
of prisms I',II',
of the volumes
III',
etc.,by S' ; and the volume of prism
S'
v'.
I' by v'. Then
s
=
"
V1
But
9.
"Sf
and
8 "
S'.
10.
.-.
V* +
S"
V+
11.
.-.
V'
V"
S'
12.
By making
"
smaller
and
hence
V'
S ; i.e.
"
of the altitude
v' may
be made
is "
than
Similarlyit
not
16.
Ex.
its rightsection
1325.
its lateral
area
Hint.
Ex.
one
that
See
1326.
The
volume
and
a
F* is
16.
of
lateral
obliqueprism
edge.
an
is
equal to
"161,6.
the
product
of
product
of
788.
volume
The
and
one
Ex.
1319.
The
diagonal
13.
r'.
"
Q.E.D.
-
Apply "
Ex.
proved
a
to 2-14.
V'.
1324.
Hint.
; i.e. V is not
be
may
V=k,
"
V.
"
r=
.-.
be made
v',may
previouslyassigned
any
constant, is false
15.
less than
small.
small.
volume, however
.-.the supposition that v'
14.
11.
V"vf.
"
smaller,prism I',
and
F, which
"
less
and
V1
previously assigned volume,
however
.*.
10.
the divisions
QR
any
13.
V.
half the
base
42
of
of
a
inches.
regularprism
a
apothem
prism
equal
to
the
of its base.
is a rhombus
If the
is
altitude
having one
of the prism
side 29 inches
is 25
inches,
find its volume.
Ex.
1327.
lengths
volmn^of
In
of
a
its
the cube.
certain
edges
cube
the
area
have
the
same
of the
surface
numerical
and
value.
the
Find
bined
com-
the
374
SOLID
GEOMETRY
Proposition
The
804.
of
Theorem
is equal
triangular pyramid
product of its base and its altitude?
the
third
one
volume
XII.
a
c
C
triangularpyramid
Given
by V, its
To
base
and
by B,
\B
F=
prove
"
1. Construct
The
its altitude
denoted
H.
is
Only
with
AG
then
base ACD
composed
quadrangularpyramid
and
O-ACD
by
its volume
H.
prism
prism
with
O-ACD
Argument
2.
and
of
lateral
edge
O-ADGF.
Through OB and OF pass a plane,intersectingADGF
and
into two
dividing quadrangularpyramid O-ADGF
pyramids
4.
iDGFisaO;
5.
.\
6.
But
base
O-ADF
and
in
D
.*.
3 times
9.
.*.
V=
O-ACD
1328.
Find
whose
edge
Ex.
then
O-ADF
O-DGF
+
OGF
D-OGF
=
O-DGF
=c=
ABC,
O to any
1329.
is the
of A
center
Find
the
the
base, as
volume
height2 V3
V to the
ABC
plane
(" 752).
ABC,
and
of A
the altitude
edge of
slant
of the _L from
of
; with
OD
a
=
\
of
a
_L
OA.
hedron
regular tetra-
altitude
a.
of
prism AG.
q.e.d.
hedron
regular tetra-
a
as
O-ACD.
volume
of 0-ACD=the
of
be taken
prism AG.
is 6.
foot
f of
=
the volume
may
=c=
=
O, the
with
ADGF.
=
i the volume of prism AG.
.".V="B-R.
prism AG
B.H;
Ex.
angular
tri-
/
the volume
But
from
;
+
10.
QA
ADF
O-DGF.
vertex
as
8.
Hence
A
in
O-DGF.
triangularpyramid O-DGF,
But
of base
.-.
o=
7.
Hint.
and
O-ABF
CO.
triangular pyramid
3.
DF
to
Proposition
805.
the
The
volume
of
of
base
product
its
375
VII
BOOK
XIII.
Theorem
and
equal
is
pyramid
any
to
third
one
altitude,
its
O
D
C
pyramid
Given
base
by
To
B, and
The
its
proof is
See
Hint.
B
left
as
Any
II.
X.
Cor.
III.
to each
are
other
inches,and
18
=
A
1331.
Ex.
the
of
pyramid and
OG
to the
prism
a
have
(b) two
other
figureof " 805, if the base
inclination
the
as
having equivalent bases
to each
are
other
altitudes.
their altitudes, and
as
In the
1330.
pyramids
altitudes
having equal
Ex.
(a) Two
their
to each
are
and
bases
equivalent
pyramids
two
products of their bases and
808.
V, its
exercise for the student.
an
I.
Cor.
by
H.
Pyramids
having
equal altitudes are equivalent.
807.
denoted
H.
-
proof of Prop.
806. Cor.
its volume
altitude,OQ, by
\
V=
prove
with
O-ACDFG
as
250
=
their
square
60",find
is
base
pyramids
bases.
inches, OO
the volume.
equivalentbases
and
equal
altitudes ; find the ratio of their volumes.
809.
Historical
is the
pyramid
altitude
of
the
"
is attributed
Athenian
212
b.c),
for
the
called
surface
Archimedes,
see
The
Note.
third part of
to
School.
Sphere
and
proof of the proposition that "every
base and with the same
prism on the same
a
Eudoxus
In
and
(408-355 B.C.),a great mathematician
written
work
a noted
by Archimedes
(287there
is
found
also
an
Cylinder,
expression
of
volume
"" 542, 896,
was-g^ven by Brahmagupta,
and
a
a
pyramid.
973.)
noted
Later
Hindoo
(For
a
a
further
solution
writer
bom
account
of this
about
of
problem
598
a.d.
376
GEOMETRY
SOLID
XIV.
Proposition
Theorem
triangular pyramids, having a trihedral
angle of one equal to a trihedral angle of the other, are to
other as the products of the edges including the
each
D
equal trihedral angles.
Two
810.
triangular pyramids
Given
Z.0
Z
trihedral
=
and
O-ACD
with
Q, and
Q-FGM
volumes
denoted
hedral
tri-
by
V
v'jrespectively.
and
To
prove
V
OA-
00-
01)
V1
QF-
QG-
QM
Argument
1.
with
pyramid
Place
Z. Q
Reasons
Q-FGM
so
shall coincide
with
Eepresent pyramid
positionby O-F'g'm'.
2.
From
a
Then
and M' draw
D
A
DJ
OAO
"
trihedral
in
Q-FGM
and M*K _L
A
DJ
trihedral
that
Z
its
1. " 54, 14.
0.
new
plane OAO.
OAO
DJ
_
"
"
=
.
V'
A
OF'G'
A
OA
OAO
"
"
i/'iT ~~A
OF'G'
M'K
00
But
OF'G'
A
A
OF1
"
OG'
by DJ
plane determined
if 'ifintersect plane OA C in line OKJ.
Again
Then
let the
rt. A
BJO
rt. A
~
and
M'KO.
OP
DJ
_
M'K~
OM''
V
OA-OC
V'
OF'
OD
OA-OC-OD
OM'
QF.QG-
8.
Q.E.D.
"
OG'
" 309.
QM
they have the
of faces similar each to each and similarlyplaced,
number
same
and have their correspondingpolyhedralangles equal.
811.
Def.
Two
polyhedrons
are
similar
if
378
SOLID
GEOMETRY
XVI.
Proposition
to
equal
of
sum
of
volume
The
815.
the
third
one
frustum
a
its two
between
frustum
To
its altitude
the
pyramid
2.
+
b+
V
B
Then
the
b).
"
minus
1.
" 54, 11.
+
of
O-RM.
2.
" 805.
3.
" 757.
4.
"
5.
Solving for
6.
" 309.
h')" "5 H1
-lH\B-b).
.
to find the
value
of H(.
if
b_
3.
O-AF
altitude
"HB
remains
now
its
Reasons
J b(h+
V=
=
It
noted
de-
O-RM.
H' denote
Let
portional
pro-
D
pyramid
=
mean
the
pyramid O-AF, with its volume
base by B, its upper
base by b, and
lower
AM
and
bases.
Argument
1. Frustum
is
pyramid
any
of
AM,
V=\H(B
prove
of
base, and
C
by V, its
altitude by H.
of
product
base, its upper
its lower
Given
Theorem
=
B
y"
4.
.
H'
=
H
5.
Whence
H*
54, 13.
+ tf'
HVb
=
VJ3-V"
VB-Vb
=
i.e.
V=\
\HB
+
H(B
\Hb
-h 6+
+
"H^B-b;
Vtf 6).q.e.d.
"
H
'.
BOOK
Proposition
A
816.
truncated
379
VII
XVII.
Theorem
-prism
triangular
triangular pyramids whose
and
vertices
whose
frustum
bases
three
the
of
inclined
the
truncated
Given
To
section.
the
are
the
are
equivalent
three
F-ACD
=c=
G-ACD
+
+
Through A, D, F
three
K-ACD.
and
of the
F-DGK
pass
planes
into
ACD-FGK
Since
F-DGK
is
F-ACD
required pyramids,it
F-ADK
and
K-ACD
o
mains
re-
G-ACD.
o
.-.the altitude
the altitude
F-ADK
But
=c=
in
and
F-ADK
.'.
ACD
A CD
as
K-CDG
=c=
"
=
may
be
taken
=
K-CDG
as
vertex.
K-ACD.
F-DGK
and
F-DGK
of
pyramid F-ADK
pyramid C-ADK.
C-ADK.
K
o
Likewise
.'.
of
C-ADK,
base
-'
F
IIplane AG.
CF
.-.
K, D,
frustum
to prove
.-.
Reasons
triangular pyramids F-ACD,
F-ADK,
one
and
FGK
'""K-ACD.
=c=
C-DGK
=0= A-CDG
;
G-ACD.
=
G-ACD.
=0*
F-ACD
+
G-^CZ)
+
Q.E.D.
1.
to
of
vertices
_-^-n(?
Argument
dividing
base
triangularprisinACD-FGK.
ACD-FGK
prove
is
"611.
380
SOLID
817.
Cor.
prism
the
and
818.
I.
The
is
equal
Cor.
volume
to
of
The
n.
truncated
a
third
one
its lateral
of
sum
GEOMETRY
the
right triangular
product of
edges.
volume
of
^/I^
into
triangular prism QR
triangularprisms.
Ex.
sides
I
jk/^
/ /
/
truncated
truncated
two
/
/V^-,Y
J--J
right Qi^---
^^"^^
of
base
The
1334.
divides
ACD
section
II
I
triangular prism is equal
to one
third the product of a right section and
the sum
of its lateral edges.
Rt.
I
-^
any
truncated
Hint.
its base
truncated
a
; its lateral
13, 14, and 15 inches
right triangularprism has
8, 11, and
edges are
for its
13 inches.
Find
its volume.
Ex.
result
Ex.
The
1337.
cubic
and
divides
the
section ;
0
=
and
compare
result
compare
altitude
with
Ex.
and
4
altitude
an
of
13
inches,respectively.
altitude of the
base
of
pyramid contains
the
into
the
base
the
frustum
cuts
squares
The
the
of the frustum
the
is the second
?
144
is
Find
inches, and
square
pyramid
parallel
:
(a)
the
base
the
to
the
its
base
of the
area
formed.
pyramid parallelto
a
of the frustum
cuts
off
a
mid
pyra-
areas
of any
of two
tetrahedrons
are
to
each
homologous edges.
two
altitude of
similar
a
pyramid
is 6 inches.
equivalentparts.
into two
A
plane parallelto
Find
the
altitude of
formed.
Two
bushels, and
of
pyramid
thus
1342.
the
equal parts.
two
total
The
1341.
of
section
A
pyramid
square
given pyramid.
the
to
a
a
frustum.
the
section
A
1340.
as
2| inches
are
has
pyramid
square
bases
(6) the volume
similar
Ex.
a
Find
inches.
1339.
Ex.
5
The
is 10
altitude
wheat
other
800
bins
are
bushels.
similar in
shape
; the
If the first is 15 feet
one
deep,
holds
1000
how
deep
ting
passed parallelto the base of a pyramid cutFind : (a) the ratio of the section
the altitude into two equal parts.
the base ; (") the ratio of the pyramid cut off to the whole
pyramid.
Ex
to
and
B
=
6
edges of the bases of a frustum of
and the volume
inches, respectively,
inches.
1338.
Ex.
Ex.
b
(1) put
volume.
3 inches
other
"815:
(2) put
;
of
edges of the
; the
the
204"
805
"
frustum
A
1336.
Find
of
of
of " 799.
Ex.
inches
formula
the
formula
with
formula
are
In
1335.
1343.
A
plane
is
BOOK
EXERCISES
MISCELLANEOUS
of
edges
faces of
the
a
the
three
if their volumes
Construct
three
to the
farmer
A
1349.
Ex.
the
has
is
will it
corn
for
ear
of
1 bushel
of
bases
a
of
bushels
in
how
of
the
lel
paral-
vertex.
section
of
If the crib
in the
corn
the
mation,
approxi-
exact
form
a
bushels
many
the
is 30 feet
volume
of
a
tum
frus-
high
edges of its
feet, respectively. How
many
12 feet
and
wheat
will it hold ?
6
the
long, a cross
denoting feet.
numbers
For
from
section
a
20 feet
(Use
corn.
feet.
elevator
wheat
pyramid
square
are
1\
1439.)
=
A
1350.
Ex.
cubic
Ex.
bushel, see
a
shelled
of
area
; 4 inches
2 bushels
contain,counting
bushel
1
the
is
Its base
inches.
is 12
Find
base
representedin the figure,the
in the ear,
entirelyfilled with corn
which
822.
corncrib
a
ratio of 1 to 27.
base.
pyramid
a
from
4 inches
and
base
of
and
equivalentpyramids on the
of all pyramids equivalent
same
side is 5 inches.
whose
regularhexagon
the
in
are
homologous edges
in the
are
vertices
the
ratio of the
the
homologous edges
more
or
821 and
Exs.
altitude
The
whose
and
ratio of their
the
Find
areas
with
Compare
1348.
Ex.
of
points equidistant from
(6) Find the locus of
given pyramid and standing on
Hint.
of
of all
base.
same
is
three
tetrahedrons
(6)
(a)
1347.
Ex.
the
ratio of the volumes
the
similar
of their total
ratio
locus
Find
(a)
of two
areas
points equidistantfrom
angle.
ratio of 2 to 5.
the
to
the
Find
1346.
Ex.
of all
angle.
trihedral
a
locus
the
trihedral
a
1345.
Ex.
total
Find
1344.
Ex.
381
VII
; the
the
(Use
approximation given
in
Ex.
1349.)
Ex.
12
1351.
inches, and
Find
Ex.
of
the
the
the
edges
In
a
a
regular square
of its bases
of the
frustum
10 and
are
of
of
4
are
pyramid
has
pyramid
inches
of which
and
10
an
altitude of
inches,
the frustum
is
a
tively.
respec-
part.
regularquadrangular pyramid, the
and the slant height is 14.
6, respectively,
a
sides
Find
volume.
Ex.
whose
Ex.
as
frustum
the volume
1352.
bases
A
1353.
Find
altitude is 8
1354.
large ;
n
lateral
inches,and
The
times
the
as
edge of
large.
a
each
cube
of
area
a
regular triangular pyramid
side of whose
is
a.
Find
the
base
is 6 inches.
edge
of
a
cube
3 times
382
SOLID
Ex.
1355.
the form
of
contains
a
bottom, and
67.2 cubic
Ex.
inches.
The
1356.
10 feet.
x
of
Ex.
deep
The
1357.
the
BD,
Find
Let
AB,
the
more
less than
or
a
6, and
given in "
777.
Would
if the
Egyptian
1359.
How
the
denoting
feet in the
barn.
in
c,
values
your
result
in Ex.
1357.
formula
Ahmes'
of coal ?
10 tons
numbers
rectangle
a
formula
Compare
result obtained
the
correct
Ahmes'
in
c
of cubic
6, and
the
the
is
and
barn
a,
coal bin
a
to hold
barn,
a
number
of the
dimensions
for
be made
bin
BC,
1357, be denoted by
respectively. Substitute
with
contain
box
basement
a
figure represents
Ex.
a,
the
must
in feet.
1358.
Ex.
left in
space
How
dimensions
of
the
Does
?
quart
8
supposed to contain a quart of berries is in
pyramid 5 inches square at the top, 4| inches
2$ inches deep. The United States djryquart
box
berry
frustum
a
the
at
square
A
GEOMETRY
have
barns
been
had
been
similar
in
shape
in Ex.
to the barn
1357?
Ex.
1 cent
per
for the
The
1360.
and
3 feet
Ex.
barn
Ex.
divide
Ex.
at the
are
load ?
Ex.
the
1357
in the
are
of
has
and
2
a
a
cents
1357
in Ex.
barn
per
at
foot
square
if the
feet of masonry
the sides of the barn.
planes as
same
18 inches
foundation
stone
of cubic
number
regular tetrahedron
of
a
is
is
octahedron
regular
dam
bottom,
across
and
there in the
Give
the
\6-V2. Find
its
that the
Prove
a.
pyramid,
is
six
is 40 feet
stream
4 feet wide
at
the top.
long, 12
How
feet
many
a
parallelo-
high, 7
feet
feet of
cubic
loads,counting 1 cubic yard to
geometrical solid representedby the dam.
? how
many
S the lateral area,
and
H
the
altitude,of
a
regular
find the volume.
Find
1366.
a
dam
by the diagonalsof
equivalentpyramids.
determined
of the
name
Given
1365.
total surface
planes
parallelopipedinto
A
1364.
material
square
the
altitude.
edge
The
1363.
piped
Ex.
paint
equals" V2.
volume
a
The
1362.
Ex.
the
volume
The
1361.
Ex.
in
Find
deep.
edge, slant height,and
wide
to
lateral surfaces
for
of the walls
surfaces
outer
foot
square
will it cost
roof ?
Ex.
wide
much
T,
and
the
volume
V, of
one
edge
of the base
a
regular square
is
a.
pyramid,
if its
VIII
BOOK
CYLINDERS
CONES
AND
CYLINDERS
819.
Def.
A
cylindrical surface
moving straightline that
and remains
parallelto a
the given curve.
CD
is
a
surface
1.
820.
continuallyintersects a fixed curve
fixed straightline not coplanar with
CylindricalSurface
Fig.
By referringto ""
Defs.
693
and
2.
Cylinder
694, the
student
give the definitions of generatrix,directrix, and element
surface.
Point these out in the figure.
cylindrical
may
The
a
a
K
Fig.
a
generated by
student
broken
should
line to
a
note
curved
of
that
by changing the directrix from
a
line,a prismaticsurface becomes
surface.
cylindrical
821.
A
dary
figurewhose bounconsists of a cylindrical
surface and two parallelplanes
cutting the generatrixin each of its positions,
as DC.
Def.
cylinder is
a
solid closed
383
384
SOLID
822.
bases
the
The
Defs.
cylinder,as AC
cylindricalsurface between
cylinder;
surface
perpendicularto
824.
not
825.
from
other
The
point
base,as HE
any
the student
(a) Any
("" 618
and
(b)Any
(c) A
a
portion of
is the lateral -surface
element
bases
in
should
of the
is
drical
cylinof
element
an
elements
are
elements
Oblique Cylinder
cylinderwhose
elements
the bases.
of
cylinderis
a
plane of one
Figs.3 and 4.
prove
4.
a
in the
followingare
two
the
cylinderwhose
Fig.
altitude
of the
some
the
of
base
to
perpendicular
the plane of the
propertiesof
correctness
a
the
cylinder
of each
are
a
cylinder;
:
parallel and
equal
634).
element
of a rightcylinderis equal to
line drawn
cylinderparallelto
element
the
oblique cylinder is
perpendicularto
The
between
an
called the
the bases.
An
Def.
826.
portion of
Right Cylinder
Def.
(Fig.4) ;
DF
the bases
right cylinder is
.
3.
the
and
are
MN.
A
Def
Fig.
and
included
as
cylinder,
823.
are
parallelplane sections
two
of the
of the
the
GEOMETRY
("" 822
and
through
an
any
element,and
179).
its altitude.
point in the lateral surfaceof
limited. by the
an
bases,is itself
a
386
SOLID
Ex.
1367.
its base
is
Ex.
a
the
Every section of a cylinder made
circle,if the base is a circle.
If
1368.
line passes
line
a
the
through
bases, if the bases
830.
by
GEOMETRY
A
Def.
joins the
circles.
of the bases
centers
of
a
'
of
right section
plane perpendicularto
a
plane parallelto
a
cylinder,this
of every section of the cylinder parallelto
center
are
by
cylinderis
a
element, as
an
section formed
a
section EF.
WB
Fig.
1.
Cylinder
Circular
Base
831.
is
A
a
base
833.
base
AB
834.
will
be
When
a
circle
a
O
?
Ex.
1370.
surface
of
Ex.
smaller
the
same
to
end.
a
a
rt. section
rt. section
and
in which
bases.
circular
exercises
the
bases
also
See
the locus of all
2 feet
right
a
is
EF
",
a
rightcylinderwhose
EF
be
a
on
the
of
O ?
a
O
the
In
?
cylinder that follow
cylindersare
be
Fig. 2, is
understood
circles.
to
mean
" 846.
points at
a
distance
of 6 inches
from
long.
the
locus of all
points : (a) 2 inches from the lateral
altitude is 12 inches and the
right circular cylinderwhose
of whose
1371.
Fig. 1, is
theorems
cases
Find
a
cylinder is
Fig. 3, would
Find
1369.
which
if rt. section
cylinder is used, therefore, it must
straightline
radius
In
The
limited
with
cular
Right CirCylinder
(Fig.3).
In
Note.
cylinder
3.
cylinder.
circular
right
Questions.
a
Fig.
cylinderin
a
thus, in Fig. 2,
circular
a
A
the term
Ex.
a
is
Def.
is
Circular Cylinder
cylinder is
circular
circle ;
cylinderAD
832.
Fig. 2.
AB
Def.
section
with
is 5 inches
base
A
log is
Find
the
size at each
20
;
feet
the entire
long
and
of
the
largestpiece of
be
cut
from
dimensions
end, that
(") 2 inches from
can
30
inches
the
in
log.
surface.
diameter
square
at
the
timber,
BOOK
II.
Proposition
Every section of
835.
through
Theorem
cylinder made
by a plane passing
is a parallelogram. (See " 834.)
a
element
an
387
VIII
E
cylinderAB
plane through
with
Given
by
a
base
element
CF
CF, but in the circumference
To
CDEF
prove
AK, and
and
CDEF
a
point,as
some
Through
2.
Then
line
the
a
it is the
4.
with
the
is
.'. DE
6.
Also
7.
.-.CDEF
836.
made
a
by
a
a
" 826, c.
1.
Arg.
"614.
DF
cides
coin-
DE.
str. line and
is
Cor.
plane DF.
of plane
surface,and
cylindrical
and
CD
" 179.
plane DF IICF.
is an element;
surface.
cylindrical
intersection
with
5.
line in
this line lies also in
But
in
Reasons
drawn
so
i.e. it lies in the
3.
D, not
O.
a
draw
D
made
of the base.
Argument
1.
section
EF
CF.
str. lines.
are
O.
is II and=
"
q.e.d.
5.
" 826, a.
6.
" 616.
7.
"240.
of a right circular
plane passing through an element
Every
section
cyliivder
is
a
is 4
inches,
tangle.
rec-
"
Ex.
element
makes
Ex.
In
1372.
CF
with
1373.
is 12
CD
the
figureof Prop. II, the
is 1 inch from
inches, CD
an
angle of
Every
How
parallelogram.
60".
section
of
Find
a
is the base of
the
radius of the base
the center
area
of the base, and
of section
CF
CDEF.
cylinder,parallelto an element, is a
this cylinderrestricted ?
(See " 834.)
388
SOLID
GEOMETRY
CONES
837.
A
Def.
conical
is
surface
surface
a
generated by
intersects
moving straightline that continually
and that passes through a fixed
point not in the plane of the
a
fixed
a
curve
curve.
838.
By referringto
dent
"" 693, 694, and 746, the stugive the definitions
may
Defs.
of generatrix, directrix, vertex,
and
element,
nappes
of
Point
these
The
that
and
upper
conical
a
in
out
student
a
surface.
the
figure.
should
observe
directrix
the
by changing
from
lower
line to
broken
line,a pyramidal surface
839.
whose
A
Def.
is
cone
boundary
Conical
curved
a
becomes
a
a
conical
solid closed
extending from its vertex
a
plane cutting all its elements, and
section formed
by this plane.
conical
surface
840.
the
822,
By referringto ""
Defs.
student
give
may
surface, and
of
out
Point
cone.
841.
section
center
Thus
0, such
842.
Def.
A
such
that
circular
in
Fig. 4, if
that
Def.
vo
The
to the
in the
is
cone
to
the
and
748
element
Fio. 2.
figure.
a
containing a
cone
joiningthe vertex
perpendicularto the
line
a
of the section is
its vertex
Fig.5\
these
a
definitions
the
of vertex, base, lateral
a
surface.
figure
portionof
consists of the
section
is J_ the
altitude
plane
of
AB
of its
a
cone
V-CD
is the
base,as
VC
of the
Cone
circular
cone
to the
section.
is
V-CD
cone
cone
section,
of
Surface
is
a
a
O with
circular
center
cone.
perpendicularfrom
in Fig. 3 and
VO in
BOOK
843.
In
Defs.
its vertex
Fig.
Circular
with
circular
a
base,if the line joining
is perpendicular to the
of its base
to the center
Cone
3.
with
cone
a
389
VIII
Base
Fig.
Cone
Circular
4.
V
plane of
the
If such
line is not
a
of the
plane
oblique
The
is the
line
of its
is called
cone
of
axis
of each
a
vertex
are
the
cone;
Fig. 5.
of
some
student
should
of a rightcircular
to
necessarilyto
must
Ex.
perimeter of its
The
Note.
limited
cases
theorems
in which
circular
be understood
2 inches
12 inches
and
the
mean
base
and
cone
a
cone
the
the
radius
is
the
prove
ness
correct-
of the
with
of all
base,
of
of whose
equal to
the
on
cones
term
are
cone
circular
points
a
of
its altitude.
to
cone
a
any
element.
an
the
equal.
are
vertex
exercises
bases
is the locus
from
is
When
cones.
to
What
1374.
surface, and
lar
Circu-
Cone
the
(b) TJie axis of a rightcircular cone
(c) A straightline drawn from the
846.
Right
:
elements
the
the
to
vo, Fig. 5.
following
properties of
point in
an
circular
right
a
joiningits
base, as
The
(a) The
the
perpendicularto
base, the
Def.
845.
be
cular
cir-
right
a
(Fig.3).
cone
844.
center
is
cone
(Fig.5).
cone
cone
base, the
cone
inches
right circular
base
will
circles,
though not
is used, therefore,it
base.
2
that follow
is 5 inches
See
also
from
cone
?
"
834.
the lateral
whose
tudes
alti-
390
SOLID
GEOMETRY
Proposition
847.
through
Given
a
section
Every
its vertex
plane through
To
Theorem
made
cone
a
by
triangle.
a
with
base
a
plane passing
V
and
AB
section
V
Then
the
by
A.
a
Argument
From
made
VCD
V.
VCD
prove
of
is
V-AB
cone
III.
Reasons
str. lines to C and
draw
D.
1.
" 54, 15.
elements
2.
" 845, c.
3.
" 603,
4.
"614.
5.
" 616.
6.
"92.
'
lines
drawn
so
are
i.e. they lie in the conical
But
.-.
lines lie also in
these
they
with
VCD
coincide
Also
.*.
of the
and
VC
the
Can
VD,
CD
are
respectively.
str. lines and
VCD
q.e.d.
kind
What
of
Find
the
locus of all
given straightline, at a
be if the given angle is 90"
a
surface, and
str. line.
a
Ex.
plane
a.
trianglein general is the section of a cone
is oblique ? if the cone
is a rightcircular
vertex, if the cone
of an oblique cone
be perpendicular to the base
any section
of a right circular cone
?
Explain.
1376.
locus
with
conical
VCD.
of
A.
cone?
Ex.
with
a
1375.
?
the
VC, VD, and
through
cone
a
plane
intersections
with
is
CD
is
Ex.
the
are
;
surface.
1377.
Find
given plane
angle is 90" ?
the
at
a
straightlines making
given point in the line.
a
given angle
What
will this
?
locus of all straight lines
given point.
What
making
a
will the locus be
given angle
if the given
BOOK
Proposition
848.
section
Every
to its base
is
of
circle.
a
IV.
a
391
VIII
Theorem
made
cone
by
a
plane parallel
by
a
plane IIbase
(See " 846.)
V
Given
To
CD
section
a
section
prove
of
CD
and
R
be any
S
two
CD-, pass planes through
2.
Prove
A
3.
Then
"".
VOM
A
~
VO
Proof
of
pointson
and
OF
and
VPR
andA
"
AB.
O.
a
Outline
1. Let
made
V-AB
cone
and
points R
A
VON
S.
OM
i.e.
"
of section
A VPS.
~
VO
ON
boundary
the
ON
=
"
"
'
PR
4.
two
But
OM
ON-,
"
the
points on
5.
.-.
section
849.
Cor.
to the
base
is to the
PS
VP
PS
Any
the
square
is
a
of section
any
CD.
O.
q.e.d.
of a cone
parallel to its base is
from the vertex
of its distance
altitude
of the cone.
section
square
of
PS
PR-, i.e. P is equidistantfrom
=
boundary
CD
as
.-.
PR
VP
the
V
Outline
section
By
" 563,
of
CD
Proof
PR
Prove
Then
base
"F$r
AB
VF2
applicationsof "" 848 and 849,see
Exs.
1380-1384,
392
SOLID
MENSURATION
GEOMETRY
OF
THE
CYLINDER
AND
CONE
Areas
850.
A
Def.
element, but
an
851.
edges
figureslie
in the
lateral faces
How
1378.
Ex.
of these
How
the bases
cylinder.
in
cylinder if its lateral
a
and
cylinder,
is circumscribed
tangent
in the
the
bases
about
a
to the
and
cylinder,
plane.
same
planes
many
of the
Before
853.
student
to
be
can
that
of the
circle
the
of
area
two
to
tangent
a
cylinder if its
the
further
be
might
important steps
more
circle.
a
it
that
In
in
?
an
of
If -two
cylinder?
restricted
850-852
bases
element.
(See " 834.)
well
the
for
the
ment
developit
development
was
:
about
regular polygon circumscribed
of a regular polygon inscribed
greater,and the area
or
less,than the area of the regular circumscribed
(1) The
circle is
cylindersin ""
proceeding
review
of the
a
the
if*it contains
the line of intersection is parallelto
planes intersect,
are
shown
all
figureslie
the two
cylinder
a
plane.
prism
are
point,of
of the
same
A
Def.
to
is inscribed
prism
elements
are
852.
other
no
A
Def.
is tangent
plane
of
area
polygon of
a
twice
as
many
(2) By repeatedlydoubling
circumscribed
and
sides, and
making
approach a
common
(3) This
scribed
in-
(" 541).
of
number
sides
of
regular
number
of
polygons of the same
the polygons always regular,their areas
limit (" 546).
common
be
in
inscribed
limit
(" 558).
(4) Finally follows
(" 559).
It will
the
sides
a
observed
is defined
theorem
the
the
for the
area
area
of the circle
of the
circle
is used
method
the same
precisely
of the cylinderand the cone.
four articles just cited with "" 854,
that
throughout the mensuration
the
Compare carefully
855, 857, and 858.
as
394
SOLID
Proposition
855.
the
By
bases
GEOMETRY
VI.
Theorem
repeatedly doubling the number
of
about,
of regular prisms circumscribed
sides
and
of
scribed
in-
in, a right circular
always
a
cylinder,and making the bases
regular polygons, their lateral areas
approach
limit.
common
the
attitude,R and r the apothems of the
bases, P and p the perimeters of the bases, and S and s the
lateral areas, respectively,of regular circumscribed
scribed
and inof sides.
number
bases have
the same
prisms whose
Let the given figurerepresent the base of the actual figure.
Given
To
of the
H
prove
bases
polygons,S
common
that
by repeatedlydoubling
of the
and
s
the
prisms,and making them
limit.
approach a common
number
of sides
always regular
Reasons
" 763.
2.
" 54, 8
3.
" 538.
4.
" 54, 1.
5.
" 399.
a.
BOOK
395
VIII
Argument
6.
s"
s
But
Reasons
the
by repeatedlydoubling
of
sides
the
of
and
prisms,
" 54, 7
7.
"
than
any
value,however
.*.
can
of
them
always
be
r
"
8.
" 586.
9.
" 587.
10.
"309.
11.
" 594.
ber
num-
bases
making
regular polygons,R
less
6.
s
=
a.
543, I.
the
be made
can
previouslyassigned
small.
less than
made
any
previouslyassigned value, however
small.
R
9.
s
be
can
less than
made
any
R
previouslyassigned value, however
small,S being a decreasingvariable.
~
10.
S
.-.
s, being
"
always equal to
S
,
R
be made
can
less than
assigned value, however
11.
.-.
S
and
approach
s
a
previously
any
small.
limit.
common
Q.E.D.
be
856.
Note.
shown
that
of its
zero
The
the
above
proof
limit of the
is limited
lateral
to
of any
area
method
prism is the same
by whatever
is successivelyincreased,provided
base
as
a
limit.
(See
also "
regular prisms, but
549.) Compare
the
that
the
inscribed
(or
of
number
each
side
proof of "
it
can
scribed)
circum-
the
sides
approaches
855
with
that
of " 546, I.
857.
common
The
Def.
limit
and
lateral
which
inscribed
the
area
of
a
right
successive
circular
lateral
areas
cylinder is the
of
scribed
circum-
regularprisms (having bases containing
of sides of the
3, 4, 5, etc., sides)approach as the number
bases is successively
increased
and each side approaches zero
as
a
limit.
396
GEOMETRY
SOLID
VII.
Proposition
The
858.
equal
lateral
of
area
Theorem
right
a
circular
product of the circumference
to the
cylinder
its base
of
is
and
its altitude.
Given
rt. circular
a
S, the circumference
To
S
prove
C
=
its lateral
cylinder with
of its base
area
of
its base
by
h.
S'
Then
P
=
"
the number
As
the
rt.
circular
regular prism.
its lateral
//.
Reasons
about
cylinder a
by
by
H.
"
Argument
Circumscribe
denoted
its altitude
C, and
by
area
" 852.
2.
" 763.
3.
" 550.
4.
" 590.
5.
" 857.
6.
Arg. 2.
7.
" 355.
Denote
by S',the perimeter
its altitude
and
by P,
H.
of
1.
sides of the
of
base
the
regular circumscribed
prism is
repeatedlydoubled, P approaches O
.-.
approaches C H as a limit.
S' approaches S as a limit.
s' is always equal to P
H.
P
"
Also
But
.'.
the
H
"
"
S=C
859.
II
limit.
a
as
"
Cor.
H.
Q.E.D.
If
S denotes
altitude, and
circular
cylinder,
B
the lateral
the radius
area,
of
the
T the
total area,
base, of
a
right
BOOK
397
VIII
S=2ttRH',
2 ttRH + 2ttR2
T=
860.
Since
Note.
the
student
of
area
equal
the
to the
element.
an
is not
with
in
of
a
This
to
since the
And
is
Find
lateral
the
dealt
only
curve
the
circle,this
been
omitted
area
and
altitude is 20 centimeters
cylinder whose
But
circular
oblique cylinder with
an
circle.
1385.
Ex.
is true.
statement
elementary geometry
and its applicationshave
theorem
oblique prism is equal
an
the
right section
base
of
area
oblique cylinder with circular bases is
tion
product of the perimeter of a rightsec-
an
and
ttR{H+ r).
perimeter of a right section and a lateral edge (" 762),
would
naturally infer that the lateral
product of
the
lateral
the
2
=
here.
total
and
of
area
a
the
right circular
of whose
diameter
base
is 10 centimeters.
open
no
allowance
for waste
In
1387.
Ex.
to the
in diameter
its total
861.
of its base
radius
the
the
a
of the
its
What
Questions.
of
right circular cylinder if
a
AD
and
BC
?
What
generate
?
might CD
be
a
generated by
as
an
axis, a
cylinder of
a
volving
rectanglereright circular
olution.
rev-
part of the cylinder will
CD, opposite the axis AB,
What
generate? What
will the plane AC
in any
one
of its
will
erate
gen-
positionsbe
f^ZZ^wT"*^
called ?
are
lateral
right circular cylinder if its lateral
base
sides
called
side
863.
ratio of the
B.
it may
of
one
sometimes
cylinderis
deep, making
its altitude is H.
Because
Def.
10 inches
?
the altitude of
is T and
area
about
862.
equal
are
Find
1389.
and
required to make
is this ratio if the altitude and
What
bases.
two
the radius
is S and
of tin will be
right circular cylinder,find
a
Find
1388.
Ex.
Ex.
inches
square
?
of the
sum
of base
the radius
area
many
cylindricalpail 8 inches
an
area
How
1386.
Ex.
Def.
Similar
cylinders
cylindersgenerated by
homologous
sides
as
axes.
of
similar
revolution
rectanglesrevolving about
398
SOLID
GEOMETRY
Proposition
864.
The
lateral
VIII.
Theorem
the
and
areas,
similar
cylinders of
squares
of their altitudes,
revolution
and
are
as
total
each
to
of
areas,
other
the squares
of
as
two
the
the radii
of their bases.
H
similar
cylindersof revolution with their lateral
denoted
areas
by S and S1,their total areas by T and T',their
altitudes by Hand
H1, and the radii of their bases by i? and R',
Given
two
respectively.
To
prove
:
N
(a)
H2
8
,
=
"
-
"
S'
IV
T*
H12
R2
=
"
-
"
R12
Reasons
1.
" 859.
2.
" 54,8
3.
" 863.
4.
" 419.
5.
" 309.
6.
" 309.
7.
" 859.
a.
399
VIII
BOOK
Argument
Reasons
R(H+R)
2L
T'~ R\H' + R')
"
+
H
Arg. 4,
H' +
R'
R
Hf
'
t
B*
"=R
11. Also
Q.E.D.
R!
Ex.
square
inches.
Find
1391.
The
7
320
square
the
largeris
Ex.
lateral
a. as
total
the
lateral
and
inches
Two
of
area
865.
(2)
The
Def.
of its base.
866.
b
Have
you
as
A
plane
867.
A
Def.
the vertex
element
A
The
869.
rightcircular
point
cones
must
in
student
cone
each
""
radius
revolution
of the
are
base
of
of the smaller.
the results in the form
?
general statement
a
to
the
if it contains
cone
a
an
cone.
in
and
cone
is
if its base
cone
a
coincides
its vertex
circumscribed
about
with
How
1393.
first is 676
is
with
cone.
pyramid
coincides
this
is inscribed
of the
is circumscribed
vertex
5
are
is a
right circular cone
to any point in the circumference
is its slant height.
of such a cone
is tangent
pyramid
of the
Def.
the
Put
axis.
proved
point,of
in the base
inscribed
868.
other
an
height of
slant
any
no
revolution
of the
area
of the base
radius
the
about
Thus
Def.
element, but
of th*
" 309.
cylindersof
similar
inches, and
straightline joiningits vertex
what
11.
a.
and
a
b; find
adjacent sides of a rectangle are
the cylindergenerated by revolvingthe rectangle: (1)
axis ;
an
total
the
of two
square
Find
general statement.
Ex.
" 309.
54,8
of the second.
area
areas
500
10 inches.
1392.
base
10.
cylindersof
similar
inches, respectively,and
and
a
" 401.
R"
altitudes of two
inches
Ex.
Rf
The
1390.
about
9.
R!
H^__H^_
H'~ H'2'
H_
10.
of
"
R'
II
H' + R'
the
8.
.
'
~
But, from
9.
H+R
J*.
._
"
"
the
the vertex
may
state
of
866-868
of
base
of the
and
correspondingto
many
about
cone
and
its
cone.
prove
the theorems
those
mentioned
be tangent
planes can
these planes pass ?
Prove.
restricted ?
the
if its
cone
a
to
(See " 846.)
a
the
in " 854.
cone?
How
on
are
Through
the bases
400
SOLID
GEOMETRY
Proposition
IX.
Theorem
870.
By repeatedly doubling the number
of sides of the
bases of regular pyramids circumscribed
scribed
inabout, and
the bases
in, a right circular cone, and
making
always regular polygons, tl%eir lateral areas
approach
limit.
common
a
Given
and
r
the
H
the
apothems
bases,and
S
and
circumscribed
same
base
To
common
s
of
of the
actual
prove
the
lateral
inscribed
and
number
of
altitude,L and
the bases,P and p
sides.
Let
areas,
heights,R
the perimeters of the
of regular
respectively,
pyramids
the
I the slant
whose
bases
have
given figure represent
of
polygons,S
and
the
figure.
that
of the bases
the
of sides
by repeatedlydoubling the number
the pyramids,and making them
always regular
limit.
s approach a common
Reasons
RL
1.
" 766.
2.
" 54, 8
3.
" 538.
4.
" 309.
5.
" 399.
a.
402
GEOMETRY
SOLID
X.
Proposition
The
873.
to
is equal
of a right circular cone
half the product of the circumference of its base
one
and
lateral
its slant
Given
rt. circular
a
C, and
by
by
-denoted
area
its slant
height by
S,
L.
L.
"
left
it the
make
to
C
Question.
as
What
for the student.
exercise
an
changes
proof of Prop.
X
in the
necessary
are
proof of Prop.
?
If s denotes the lateral area, T the total area,
the slant height, and
B the radius
of the base, of a right
875.
L
\
its lateral
with
cone
of its base
=
proof is
874.
VII
S
prove
The
area
height.
the circumference
To
Theorem
Cor.
circular
cone,
s
=
T
Ex.
1394.
radius
of the
the
The
base
=
ttRL
;
ttRL
+
altitude of
7T
i?2 =
lateral
the
#).
+
right circular
a
Find
8 inches.
tR(L
is 12 inches
cone
and
area
and
the total
the
of
area
cone.
Ex.
1395.
to make
goods
How
conical
a
is allowed
Ex.
1396.
inches,and
876.
the radius
Def.
is sometimes
similar
Def.
one
called
Similar
area
of the
Because
revolvingabout
877.
lateral
of
base
a
cone
cones
right circular
be
of its sides
a
20 feet in
is 10 inches.
it may
of
wide
30 inches
canvas
16 feet
high and
?
cutting and fitting
tent
for
The
yards of
many
diameter,
an
is
cone
Find
if
^4o
required
10% of
vgg
the
SqUare
the altitude.
right triangle
axis,a rightcircular cone
generatedby
as
will be
a
revolution.
of revolution
are
cones
generatedby
righttriangles
revolvingabout homologous sides
as
axes.
BOOK
XI.
Proposition
878.
lateral
The
of
cones
revolution
altitudes,
their
and
the
as
similar
altitudes
the radii
To
by
Apply
a
of
a
? the
of
881.
cone
Ex.
a
is the
)
as
lateral
altitude ?
The
passing through
an
.
H"
Lf%
RI2'
exercise
of
for the student.
proof used
of
a
in
Prop. VIII.
is the
cone
a
and
are
portion of
section of the
kind
of
(" 848)
?
height of
a
cone
of that
section
and
the upper
What
a
the
made
cone
cone
by
figureis
bases
the upper
of
a
a
frustum
base
of
a
rightcircular
element
an
of
of the
cone
of the frustum.
of
element, is
lower
frustum
portion of
the bases
Every
"
Tl
slant
length
1397.
R2
=
"
t"I2
What
between
areas
t/2
an
right circular
Def.
included
their
with
the base.
Questions.
cone
bases.
rW2
ir
the base
to
plane parallel
frustum
their
L2
=
-
"
of
frustum
between
880.
of
S'
method
the
A
Def.
included
left
=
"
/
proof is
879.
slant
total
H2
S
, N
(a)
V
\
Hint.
S',their
and
S
:
prove
The
their
revolution
of
cones
of
heights,
squares
of
radii
the
the
as
lar
simi-
two
by T and T1,their
by H and H\ their slant heights by L and Lf,and
of their bases by R and R',respectively.
denoted
areas
squares
of
squares
two
Given
the
of
areas,
other
to each
are
as
Theorem
the total
and
areas,
403
VIII
a
a
frustum
trapezoid.
of
a
cone,
made
hy
a
plane
404
SOLID
GEOMETRY
Proposition
882.
The
lateral
XII.
of a frustum of a right circular
half the product of the sum
of the
area
is
equal to one
circumferences of its bases
cone
Theorem
its slant
and
height.
0
frustum
Given
lateral
C
of
AM,
denoted
area
right
by S, the
c, the radii of its bases
and
circular
O-AF,
cone
R
and
its
of its bases
circumferences
by
with
its slant
r, and
by
height
byz.
To
S
prove
i(C+c)L.
=
Reasons
Argument
1.5
lateral
=
lateral
2. Let
V
of
area
of
area
denote
slant
=
It
remains
now
.
heightof cone
S="c(L + L') "CL'
Then
0 -DM.
minus
" 54, 11.
O-DM.
cone
the
O-AF
cone
$CL
+
2.
" 873.
|2/(C-c).
to find the value
of l!
.
3. " 556.
c
But
"
r
A
OKD
A
~
"422.
OQA.
R_OA_L
+ L'
r~
OD~
L*
c
L-\-L'
" 424, 2.
"
.
54, 1.
L'
c
L':
Solving for L\
CL
C-c
S
=
"CL
+
"
cL
((7-c)=l(C+c)X.
C"c
Q.E.D.
" 309.
BOOK
883.
the
L
area,
bases, of
If
I.
Cor.
a
denotes
s
405
VIII
the
lateral
the
T
area,
height, and R and r the
frustum, of a right circular cone,
S
ttL(R+ r);
T
+ r2).
ttL(R+ r)+ tt^R2
slant
radii
total
the
of
=
=
884.
circular
of
area
the
to
equal
is
cone
circumference of
the
and
lateral
The
n.
Cor.
frustum
a
product of
section
a
of
right
a
its slant
height
between
midway
its
bases.
right circular
In
formulas
results with
formulas
with
of
1400.
Ex.
inches, and
16
The
altitude
the
diameters
2
inches, OB
and
paper
its lateral
A
1402.
Ex.
whose
altitude
inches
and
are
radius
of base
Ex.
and
area
in
inscribed
find its
by "
Ex.
1406.
circular
cone
a
right circular
cone
inches
inches,
and
30
is
area.
of
arcs
Cut
circles ; OA
figure ABCB
=
of
out
area.
radii
cone
4
of bases
as
dicated
in-
of the circle
of the total
sum
inches,and
similar
of two
areas
altitude is
one
right
circular
Find
out
of
Form
area.
Do
cylindersof
f of the other.
Find
olution
rev-
the
the two
slant
a
cone
cone
altitude
difference
the
paper
regular hexangular pyramid
a
with
a
between
semicircle
with
this
results agree
whose
semicircle
20
inches
lateral
their
radius
and
and
is 4
with
areas.
inches,
find its lateral
?
height,and the diameter
Find the total
each equal to L.
The
are
results
compare
Find
cone.
a
regular triangular and
Cut
875.
compare
cut.
4 inches.
1405.
of
cylinder.
A
1404.
Ex.
120".
the radius
Find
be
square
of each
area
are
right circular
a
and
inches
1401.
The
is 216
total
is 4
1403.
Ex.
of
20
CD
ZBOC=
of
frustum
it must
which
and
also its total
and
frustum
a
a
are
inches, respectively,is made
7
in Ex.
from
and
area
also its total
and
area
C
frustum
a
0 and
r=
B
=
of
of its bases
inches, and
it into
form
b
frustum
a
figure, AB
the
5
=
bases,of
make
(a)
:
make
(")
;
of
its lateral
In
1401.
of " 883
875
"
the
height,and
" 859.
respectively. Find
Ex.
of
slant
the
CL.
=
formulas
the
L
between
midway
S
then
cone,
1399.
Ex.
section
a
lateral area,
the
denotes
of
circumference
the
a
If 8
1398.
Ex.
of the
area.
base, of
a
right
406
SOLID
GEOMETRY
VOLUMES
Proposition
I.
885.
The
polygon
and
greater
than
volume
of
which
whose, base
XIII.
Theorem
whose
prism
a
is circumscribed
about
the
volume
of the
a
regular
polygon
is
base is
a
regular
cylinder is
a
circumscribed
with
prism
twice
as
many
sides.
II.
The
and
volume
which
volume
of
polygon
with
The
is inscribed
the
inscribed
twice
as
a
their
Given
same
the
V
inscribed
number
of the actual
and
the
approach
common
v
the
prisms
of sides.
figure.
less than
whose
is
base
exercises
as
XIV.
making
volumes
H
bases,and
and
left
a
the
regular
for the student.
Theorem
the number
circumscribed
cylinder,and
regular polygon
a
cylinder is
a
By repeatedly doubling
of prisms
base is
sides.
many
Proposition
bases
in
prism
figuresand proofs are
886.
whose
of a prism
about,
bases
a
and
sides
of
inscribed
and
always
common
altitude,B
of
the
in,
regular polygons,
limit.
b the
areas
of
the
of circumscribed
volumes, respectively,
whose
regular and have
given figurerepresent the
bases
Let the
are
the
base
BOOK
To
of sides of
by repeatedlydoubling the number
the prisms,and making them
always regularpolygons,
limit.
v approach a common
that
prove
the bases
of
and
V
40T
VIII
Reasons
Argument
V=
B
v
and
H
"
"B
v
H
'
b
=
1. " 799.
H.
-
B
.
2.
"
3.
" 399.
4.
" 54,7
"
"
v~
b
V
H~
"
B
v
"
b'
54,8
a.
b
"
V
V"V
V
=
a.
5.
Similar
to
Arg. 7, "
855.
5.
" 546, II.
6.
Similar
to
Arg. 8, "
855.
6.
" 586.
7.
Similar
to
Arg. 9, "
855.
7.
" 587.
8.
Similar
to
Arg. 10, "
8.
" 309.
9.
.*.
9.
" 594.
same
manner
and
V
855.
approach a
v
limit.
common
Q.E.D.
887.
proof of "
Ex.
a
Read
855.
is 280
Ex.
Ex.
from
and
.
must
the
which
Def.
the
a
be
third ?
one
volume
successive
prisms approach
as
a
total
from
off
by
a
altitude is
whose
cone
of the
area
cut
cone
off
hy
the base.
right circular
cone
is 12 inches.
the
plane parallelto
rightcircular
passed parallelto
of the
area
The
the
6 inches
is cut
altitude of
plane
one
the
as
What
base
and
?
vertex
total
?
originalcone
888.
altitude of
The
1409
the vertex
area
the
in the
right circular
a
Find
and
base
lateral surface
part of the
6 inches
the
The
1408.
of
area
inches.
square
plane parallelto
886 is limited
" 856.
total
The
1407.
10 inches
proof of "
The
Note.
a
cut
cone
increased,and
the
base
off shall be
so
one
far from
How
the
lateral
half that
of the
that
nth ?
of
a
volumes
the
is II.
cone
cylinder is the
of circumscribed
number
each
side
of
sides of the
approaches zero
limit
common
and
inscribed
bases
as
a
is
cessively
suc-
limit.
408
SOLID
GEOMETRY
Proposition
The
of
volume
its base
Given
To
cylinder is equal
a
its volume
cylinder,with
a
V=B
prove
Theorem
to the
by
"
denoted
by
Circumscribe
base
Denote
the
2.
Then
3.
As
V'
the
=
B'
number
the
of
its
Reasons
the
cylinder a
is a regular polygon.
volume
by V* and
of its base
area
.
a
B'
of sides
5.
Also
v'
6.
But
Vf is
7.
.*.
R the
of the
base
3.
"558.
4.
" 590.
5.
" 888.
6.
Arg.
7.
" 355.
peatedly
prism is reapproaches B
limit.
.-.
890.
" 852.
2. " 799.
H.
circumscribed
4.
"
1.
B'.
by
doubled, B'
as
by
H.
about
prism whose
V, its base
H.
Argument
1.
product
its altitude.
and
its altitude
B, and
of
XV.
approachesB
H
approaches
V"B
"
Cor.
radius
V
"
as
H
as
a
always equal to
limit.
a
limit.
B'
"
H.
H.
Q.E.D.
2.
// v denotes the volume, H the altitude, and
of the base, of a cylinder,
V=*R2H.
410
with
base
a
is 8 times
The
1412.
Ex.
rectangle: (a)
the
Compare
of
dimensions
Find
respectively.
of the
that
volume
the
its
about
the
of
altitude
an
total
of
area
inches
12
similar
a
and
cylinder
given cylinder.
of
solid
the
volumes
with
8
inches,
generated by revolving
axis
an
as
and
6 inches
rectangle are
a
longer side
of these
ratio
has
Find
of 5 inches.
radius
volume
whose
cylinder of revolution
A
1411.
Ex.
a
GEOMETRY
SOLID
(b)
;
of
ratio
the
side.
its shorter
about
of
sides
the
the
the
rectangle.
Ex.
equal altitudes
and
equal bases
Cylinders having
1413.
are
equivalent.
Ex.
1414.
their bases
Ex.
as
Any
and
each
to
are
other
products of
the
as
their altitudes.
1415.
their
cylinders
two
cylinders having equal
Two
(a)
altitudes,and
bases
(6) having equal altitudes
each
to
are
other
each
to
are
other
as
their bases.
Ex.
The
1416.
product of its lateral
Ex.
this into
Ex.
20
A
cubic
a
Ex.
Ex.
the
volume
the
Roll
inches.
6x9
paper
(two answers).
right circular cylinder is to be
will it cost to dig it at
much
a
How
in diameter.
the
altitude
of its base
In
a
the
(6) Find
of
to
of its base.
radius
find its volume
form
equal
is
?
Find
have
in the
feet
8
radius
1420.
base.
half the
one
cylinder
rectangular piece of
a
cistern
foot
1419.
is V and
out
and
deep
5 cents
and
area
right circular
a
right circular cylinder and
a
1418.
feet
the
Cut
1417.
of
volume
of
a
R.
right circular cylinder
certain
numerical
same
right circular cylinder if its volume
the volume
if the
value,
altitude
(a) Find
is
equal
lateral
the
and
radius
of the
diameter
of the
the
to the
area
base.
1421.
Ex.
Find
:
(a)
Ex.
cylinder is inscribed
the volume
1422.
diameter
A
A
quart, i.e. *f* cubic
The
(a)
State
to those
is 4 inches.
inches
student
and
given
; (ft)the
cylindricaltin
of its base
892.
of each
ratio
tomato
Does
whose
of
can
it hold
is
edge
is 10
the
cylinder to the
4T%
inches
more
or
inches.
cube.
high, and the
less than a liquid
:
the theorems
in "" 885
(6) State, by
cube
a
?
may
prove
in
aid of "
and
on
the
cone
corresponding
886.
888, the
definition
of the volume
of
a
BOOK
XVII.
Proposition
volume
The
893.
product of
Given
To
denoted
exercise
an
as
changes
the
V, its base
by
B,
in the
proof of Prop.
?
the altitude, and
H
Cor.
the
by
be made
must
If v denotes the volume,
radius
of the base of a cone,
895.
third
one
for the student.
XVII
proof of Prop.
it the
to
H.
"
What
Question.
make
to
equal
H.
V=\B
proof is left
894.
is
cone
a
its volume
prove
The
XV
by
Theorem
its altitude.
and
with
its altitude
and
R
its base
cone
a
of
411
VIII
V=\ttE*H.
Ex.
bases
and
Ex.
Ex.
of
Ex.
The
The
A
Find
the
and
that
later by
give*n
the
a
a
radius
makes
with
has
cone
long and
as
of 12 inches
the
base
an
;
an
element
angle of
30".
cone.
hypotenuse
of
right triangleis 17 inches and one
of the solid generated by revolvingthe
a
the volume
Find
side
and
cone
a
as
axis.
an
cylinder have
equal bases
and
tudes.
equal alti-
ratio of their volumes.
Historical
with
of
its shortest
1427.
proposition
bage
are
cone.
of the
volume
triangleabout
an
inches
side is 15 inches.
896.
products of their
is 18 inches and
height of a right circular cone
angle of 60" ; the radius of the base is 8 inches.
base
is 24
1426.
Ex.
the
slant
of the
volume
cone
the
other
cones
the base
1425.
the
Find
The
with
the
to each
two
altitudes.
1424.
makes
Find
Any
1423.
To
Note.
"every
same
Archimedes
cone
is the
altitude."
and
third
proof of the
cylinder on the same
is credited
Eudoxus
part of
Proofs
Brahmagupta.
a
the
also
propositionwere
(Compare with " 809.)
of this
412
SOLID
GEOMETRY
Proposition
The
897.
to each
are
cubes
their
of
volumes
of
other
the
their
of
denoted
by
slant
similar
two
cubes
of revolution
cones
their
of
heights,and
similar
altitudes,
the
as
L\
and
L
of revolution
cones
V1,their
and
V
heights by
as
Theorem
cubes
the
as
of the
dii
ra-
bases.
two
Given
XVIII.
altitudes
the
and
by
their volumes
with
their
H\
and
H
radii of their bases
by
R
slant
and
R\
respectively.
H3
V
R*
L3
"
To
prove
The
proof is left
Hint.
Ex.
of
Apply
Ex.
of
the
entire
cut
cone
other
as
given
cone
by
the cubes
Hint.
of
See Ex.
definition ?
that
Compare
the
the
Compare
areas.
fourths
total
areas
area
of
given
the
the
the
lower
X
cone
two
plane.
Prove
of the
distances
are
upper
that
planes
is
1T\ times
? of their
planes
off by the
cut
cone
cone
of their volumes
is the ratio
a
right circular
a
from
that
altitudes ?
passed parallelto
Z
plane, and
Y and
Z
the
are
vertex
to
the
each
of the
1381.
1431.
Show
Ex.
1432.
The
and
by
is three
X.
Ex.
inches
lateral
denote
off
revolution
fact follows
; their
Prop. XVI.
volumes.
Through
; let Y
base
in
of
cone
a
_.
for the student.
proof used
bases
two
what
cone,
1430.
other
; their
If the
1429.
Ex.
of the
cones
similar
a
of
method
what
cone,
circumferences
of the two
the
=
_
exercise
an
as
=
_
If the altitude of
1428.
similar
a
=
-
the
total
that
"
lateral
area
897 is
area
240 square
a
specialcase
of
a
cone
inches.
of
of Ex.
1430.
revolution
Find
is 144
the volume.
square
BOOK
XIX.
Proposition
of
volume
The
898.
Theorem
of
frustum
a
a
of its altitude
product
third
the
lower
base, its,upper
equal
the
and
the
base, and
is
cone
to
one
its
of
sum
proportional
mean
bases.
its two
between
413
VIII
O
F, its lower
by
altitude
To
by
proofis
Hint.
In the
the
of
How
high
Ex.
and
4
cubic
Ex.
circular
1436.
cone
" 815, change "pyramid"
A
pail is
base
pail be
the
Find
Find
whose
bases
of
a
of
to hold
6 feet
vertex
the
the
; the
volume
is cut
volume
by
equal
base
cone
frustum
a
its lower
a
water?
of
as
a
cone,
indicated
of
a
in
(See
frustum
; the
cone
inches.
10
base
Ex.
plane parallelto
of the
of the entire
ratio of the
altitude is
of
2| gallonsof
high
the altitude, and
H
its volume.
inches, of
is 12
"cone."
frustum
a
Find
form
in the
to
right circular
dimensions.
same
cone
the volume,
frustum
a
tin
the
feet from
of
of the
A
1435.
inches.
its
for the student.
an
radii
its upper
must
exercise
as
denotes
of the
1434.
Ex.
diameter
by b, and
base
upper
denoted
V^6).
b +
+
v
Make
1401, and
Ex.
.
If
1433.
Ex.
left
proof
Cor.
r
its
by B,
V=\H(B
The
and
base
its volume
with
O-AF,
cone
H.
prove
899.
R
AM, of
frustum
Given
1422.)
the
formed
base
is 456
cone.
to
the
to the diameter
lateral
area
of its base.
of
a
right
414
SOLID
GEOMETRY
MISCELLANEOUS
The
1437.
Ex.
base
of
whose
is
edge is 10 inches.
equal to the volume of
Ex.
1438.
long and
feet 6
2
Find
inches
the volume
and
The
1439.
Ex.
of
United
States
Ex.
vessel
a
volume
the
Ex.
from
cut
In
inside
volume
the
Ex.
is
cone
Ex.
At
base
cut the
Ex.
and
from
Ex.
the
cuts
Find
Ex.
Ex.
the volume
cone
raised
was
form
was
of
a
placing
10| inches.
water,
the
tiling(i.e. a tiling
In
used.
was
draining
2-inch
a
made
error
laid.
?
10
are
another
tilingwas
frustum
a
will it
10
side
as
side
the
volume
1444
about
its
The
slant
Find
whose
of the frustum
must
of
cular
right cirfeet and 5 feet,
a
hold, counting \\ cubic
a
circular
plane
base
is 16 inches.
passed parallelto the
be
?
angle of 120" are 10
ing
of the solid generated by revolv-
triangleincluding an
axis.
an
of the
20
as
an
of the
4tt
a
axis.
solid
as
generated by revolving the
an
axis.
rightcircular
inches.
height is
generated by revolving the
solid
longest side
height of
of its base
slant
with
cone
the volume
about
1444
circumference
a
Find
a
volume
1447.
off
was
in the
On
8 inches.
4-inch
a
equivalent parts
the
1446.
triangleof
the
diameter
used
in the
was
was
water
pond
was
a
vertex
sides of
Two
1445.
Ex.
in internal
vessel
the
of wheat
of
into two
triangleabout
triangleof
bushels
the
20, respectively.
Ex.
England,
irregularbody, it
an
base
much
as
What
altitude
cone
1444.
the
of
The
inches)
half
pond.
many
The
distance
to
of
?
1443.
what
in
the bushel
as
grain elevator in the form of
25 feet high ; the radii of its bases
bushel
a
was
the
certain
a
4
contain
to
respectively; how
feet to
volume
the
it.
A
1442.
is 10 feet
irregularsolid.
draining
drain
not
Find
volume
same
containing water.
diameter
pond, supposed
It could
regularprism
a
cylinder 18| inches
Is this the
of the
1441.
whose
of
base.
rightcircular cylinder the radius of whose
the body in the cylinder,the surface
of
Find
cylinder if its volume
bushel, formerly used
determine
To
in
cube
-a
(2150.42 cubic inches)?
1440.
immersed
be
circular
depth.
the
at
square
Winchester
right
a
in
8 inches
form
in the
can
face of
a
'
cube.
largestcylindricalpillarthat
altitude of the
the
block of marble
A
in
cylinder is inscribed
a
the
EXERCISES
A
8 inches.
remaining.
cone
is 20
inches, and
plane parallelto
Find
the
lateral
the
area
base
and
BOOK
A
Ex.
1448.
radius
is 8.16
has
cone
feet ; the
A
1449.
Ex.
in diameter
is 3 feet
feet and
a
the
volume
same
the altitude of the
feet
long
at
12.5
cylinderhaving
a
Find
the
at
diameter
4 feet in
and
of
feet.
20
log
of
altitude
an
base
of 6.25
radius
a
has
cone
415
VIII
base
whose
the
as
cylinder.
top end
the
butt
end.
How
(a)
wood
How
timber
that
(c)
size
How
In
In
of the
there
the
log ?
feet in the
whole
(c)
The
a
the
a
piece of
the
board
has
and
and
What
is the
vertex, through the
side 8 inches
Ex.
; find the
its
about
of the
area
hypotenuse
A
1452.
is 8 inches
cone
hypotenuse of
The
tin
deep
1453.
Ex.
from
the
vertex
cone
into
four
Hint.
Ex.
See
; the
areas
231
must
the
center
1454.
form
of
of
gallons
inches.)
a
cone
a
section
A
base, and
20
are
pendicular
per-
inches
cone.
inches
10
and
one
revolving the
of
is 12
of
frustum
its bases
many
be
end ?
other
sides
generated by
cubic
planes
smaller
is the
of the
righttriangleis
surface
diameters
equivalentparts
Ex.
a
thick ?
angle
tri-
axis.
an
altitude of
The
triangle,whose
Ex.
as
pail in
inches, respectively. How
liquid gallon contains
same
(c) contain, a
of 16 inches.
base, is a triangle two of whose
respectively. Find the volume of the
1451.
Ex.
the
1 inch
What
radius
a
in
ABCD.
EFGH.
cone
timber
'to the
24 inches,
timber
square
1 foot square
is square
square
of
base
through
cone
is
end
one
largestpiece of square
the
largestpiece of
feet does
larger end
the
in
length ?
board
(")
1450.
Ex.
are
being equivalent to
Hint.
?
from
cut
its
many
foot
board
feet
cubic
many
throughout
(d)
end
be
can
How
?
cubic
many
of
feet
log contain
the
does
(b)
cubic
many
are
8^
inches
and
will it hold ?
water
inches.
passed parallelto
right circular
a
At
the
what
base
to
10|
(One
distances
divide
the
?
1430.
Find
the
side is a,
volume
of
the
revolvingabout
solid
one
generated by
of its sides
as
an
equilateral
an
axis.
scribed
Regular hexagonal prisms are inscribed in and circumabout
a right circular
cylinder. Find (a) the ratio of the lateral
of the three solids ; (6) the ratio of their total areas
; (c) the ratio
1455.
volumes.
of'tfreir
416
SOLID
Ex.
How
1456.
be made
can
long and
horizontal
filled with
and
Hint.
Ex.
the
iron
Find
is
of
fo
?
inch
an
in diameter
.
right circular cylinder is 5 feet
If placed so that its axis is
inches.
of
a
gasoline to
depth of
a
inches, how
12
many
?
1024.
See Ex.
1458.
platinum
is 8
gallonsof gasoline will it contain
wire
platinum
in the form
of its base
radius
the
of
foot of
1 cubic
tank
A
1457.
Ex.
miles
many
from
GEOMETRY
and
| inch thick
pounds of
in
weight
the
the
iron
an
diameter
outer
pipe 10 feet long, if
of the
4 inches.
pipe is
(1 cubic foot of bar iron weighs 7780 ounces.)
Ex.
In
1459.
of base
equal, the total
are
value.
Find
base.
slant
Their
altitudes is 13.
Ex.
right circular
section
three
cone
midway
Ex.
far
volume
?
a
"
its
A
1465.
diagonals as
generated. Can
a
cube
about
Hint.
Make
through
Ex.
1466.
rightcircular
Ex.
circular
one
1467.
two
whose
axis.
of its
a
of
base
the
from
the
from
the
of
a
area
leaves
cone
is this
vertex
off half
by
cone
?
original
cone
a
spring
in the
the
a
plane parallel
?
why
How
is
an
V the
Find
the
the
the
total
volume.
is 2
is revolved
the
and
solid
size from
lateral
area
house, a distance of
How
inches.
volume
about
of
of
one
solid
the
generated by revolving
axis ?
volume, and
the
a
?
pipe
surface
of
to
diameter
is 6 inches
of convenient
cylinder.
cylinder. Find
the
cylinder.
a
pasteboard, pass
and
diagonally opposite vertices,
Given
a
if it cuts
plane
inside
the volume
diagonals as
cube
Given
of
rightcircular
right circular
a
to
a
far
is the
side
Find
rind
you
that
frustum
?
square
an
a
L
How
| mile, in a cylindricalpipe whose
contained
are
gallons of water
many
Ex.
of their
sum
.
vertex
is carried
Water
of
bases
upper
"
the
necessarilysimilar
1464.
Ex.
is Zl
cut from
cone
cylinder? why
and
questionsfor
same
Is every
to its base
it with
the
the
Answer
1463.
Ex.
lower
the
common
a
base.
common
volume.
cone's
from
the
R', respectively. Show
them
cal
numeri-
same
opposite sides of
lie on
and
5, respectively,
of
plane parallelto
of the
How
and
R
are
A
fourths
radius
the
between
1462.
plane ?
heights are
the
have
radius
cone.
12 and
radii of the
The
1461.
of the
altitude and
whose
cone
the volume
of revolution
cones
Find
and
surface
the volume
Two
1460.
Ex.
right circular
certain
a
R
area
T, and
the
revolve
radius
and
the
total
the cube
of
the
a
pin
hat-
rapidly.
base, of
a
area.
altitude
H, of
a
right
418
SOLID
GEOMETRY
I.
Proposition
903.
section
Every
of
Theorem
sphere made
a
by
is
plane
a
circle.
AMBN
Given
To
section
a
section
prove
of
AMBN
sphere
0
made
by
a
plane.
O.
a
Argument
1. From
0
Join
2.
Q
draw
to
and
the
perimeterof
OC
and
B, any
two
section
AMBN.
and
OQC
OQB,
OQ
4.
=
OC
5.
.'.
A
6.
.'.
QC
OQC
A
=
QB
=
-,
i.e. any
from
.'.
904.
section
which
the
a
which
center
a
OQ.
3.
By
OB.
4.
" 902,
5.
"211.
6.
" 110.
Draw
of the
not
"
AMBN
are
O.
the
iden.
a.
distant
equi-
small
circle
by
a
of
a
sphere,as
S^m
plane
the center
of
circle
by
a
of
a
plane
through the
O
AMBN.
"--
^
A
c/-"-""
made
pass
276.
q.e.d.
CRBS.
section
does
made
through
A
Def.
a
great
section
passes
sphere is
" 54, 15.
points on
two
section
is
AMBN
sphere,as O
905.
2.
points on
Q.
A
Def.
sphere is
" 639.
OQB.
perimeterof
7.
1.
AMBN.
OB.
In rt. A
3.
JL section
OQ
C
Reasons
ffjl
\^****^J"_
\
"
a
BOOK
906.
The
Def.
axis
is
sphere which
the
907.
The
Def.
of
of
circle
a
419
IX
the
perpendicularto
of
poles
circle
a
sphere is the diameter
a
of
plane of
sphere
a
of
the circle.
the extremities
are
of the axis of the circle.
kind
Considering the earth as a sphere, what
the
are
parallelsof latitude ? the equator ? the meridian
of latitude ?
is the axis of the equator ? of the parallels
poles of the equator ? of the parallelsof latitude ?
Ex.
1472.
section made
1474.
Ex.
is 784
center
is 576
the
of
of
of
area
section
a
should
with
Compare
"
conversely. (Hint.
See
(d)
The
the
of
center
axis
(e)Any
of
the
a
sphere 7
from
of the
in the
of the
radius
8 inches
the
of
area
a
from
the
sphere.
from
inches
the
center
the center.
propertiesof
of each
a
sphere;
:
sections
sphere,if two
same
in the
the center, and
are
conversely.
sphere,if two
same
is at the less distance
See
(c)In equal spheres,or
'
of
some
unequal, the greater section
equal. (Hint.
Find
are
307.
(b) In equal spheres,or
and
the
equally distant from
equal,theyare
What
What
center.
the correctness
prove
circles ?
of.a sphere 45 centimeters
section
a
17 inches.
the
Find
centimeters.
or
(a) In equal sjjheres,
Hint.
from
section
a
following are
student
sphere is
a
8 inches
area
the
The
908.
area
The
Find
ir.
plane
The
square
ir
1475.
Ex.
a
by
of
radius
The
1473.
Ex.
of circles
sections
are
the center,
from
""308, 310.)
in the
sphere,all great
same
circles
are
" 279, c.)
a
small
and
circle,
circle
of
sphere passes
a
through the
conversely.
great circles of a sphere bisect each other.
two
(/) Every great
circle
of
a
sphere
the
bisects
surface and
the
sphere.
(g) Through
of
a
diameter, there
(h) Through
and
one
909.
of
a
three
only one
Def.
surface of
exists
points on
the
and
one
only one
surfaceof
the
sphere,not
a
a
tremities
ex-
great circle.
sphere there
exists
circle
The
sphere is the
joiningthem.
the
points on
two
distance
length of
between
the
two
minor
points
arc
on
of the
the
surface
great circle
420
SOLID
GEOMETRY
Proposition
910.
All
sphere
points
II.
Theorem
the
circumference of a circle of
equidistant from either pole of the circle.
are
on
a
=^Q
Given
P
and
To
and
C
the
R
O
PC=
arc
proof is left
The
Hint.
and
circumference,
AMBK.
and
PD
RC
arc
exercise
an
as
the
points on
two
any
poles,of
arc
prove
D
RD.
arc
=
for the student.
Apply " 298.
911.
Def.
distance
pole of
The
polar
between
distance
circle
a
of
a
its circumference
point on
any
of
sphere is the
and
the
nearer
the circle.
912.
Cor.
I.
The
II.
Jn
polar
of
distance
great circle is
a
a
quadrant.
913.
the
polar
Ex.
at
a
Cor.
distances
1476.
quadrant's distance
the equator ? from
1477.
miles, calculate
the
surface
of
1478.
on
the
the equator
Considering the earth
State
the
polar
; (c) the
a
pole ?
equator
equal.
are
points on
distance
a
of
the south
pole ? from
(a)
of 23
the north
sphere
:
of the earth
distance
a
? 180" from
as
surface
the
from
? at
sphere,
same
with
the
a
radius
Arctic
a
\" from
pole ?
of 4000
Circle ;
(6)
equator.
postulate for the construction
of
a
circle
sphere corresponding to " 122, the postulatefor the
circle in a plane.
of
a
the north
point P
in miles
of Cancer
Tropic
Ex.
a
from
pole ? 23|"from
the south
Ex.
of equal circles
is the locus of all
What
the
in
equal spheres,or
on
the
tion
construc-
BOOK
Proposition
914.
a
A
-point
the
on
the
III.
Theorem
surface of a sphere
ities
of tivo other points {not the extremdiameter)
the surface, is the pole of
on
same
great circle passing through these
the
ACDB
To
and
PC
points.
two
quadrants of great
a great O
passing through points C
P the pole of great O ACDB.
prove
Given
of
"
PD
and
sphere 0,
OC, OD, and
PC
90" and
=
.'.A
and
POC
and
PD
1. " 54, 15.
90".
POD
A\
rt.
are
i.e. OP
J_
OC
_L the
plane of
.-. OP
is the
axis of O
is the
O
ACDB.
ACDB.
pole of great O
ACDB.
q.e.d.
Assuming the chord of a quadrant of
with compasses
an
arc
sphere to be given, construct
the
of
the
surface
two
through
given points on
sphere.
Ex.
Ez.
1479.
passes
a
and
given point
Ex.
at
the
and
1483.
and
given*points
at
Find
also
the
great circles
two
poles of
locus of all
given distance
a
Find
1482.
planesof
through
Find
1481.
given points
Ex.
If the
1480.
other, each
Ex.
2.
By hyp
3.
" 358.
4.
" 622.
5.
" 906.
6.
" 907.
OD
.'. OP
.'. P
Reasons
OP.
=
and
D.
Argument
Draw
of
at the distance
each
quadrant from
of
421
IX
the locus of all
a
distance
the
m
locus of
the
a
space
third
points in space
from
a
two
of
a
circle of
perpendicularto
each
equidistantfrom
given point.
two
at
a
a
great circle
distance
d
from
given plane.
all points in space
equidistantfrom
great
other.
points in
d from
are
a
equidistantfrom
given parallellines.
two
422
SOLID
GEOMETRY
Proposition
intersection
The
915.
of
circwrriference
two
Given
To
IV.
prove
of
circle.
a
of
0.
a
Outline
1.
Show
that
2.
Show
that AC,
3.
Show
that
4.
The
5.
.*.
given point
16
is the intersection
916.
locus
10 inches
is the
of
area
from
angle formed
arcs
917.
angle
AB
and
formed
by
of great circles of
*
A
advanced
different
meaning
mathematics.
O.
a
(" 614)?
is
line
cumferen
is the cir-
in
Z
plane A
a
12
is 24
centers
inches
inches.
and
Find
circle of intersection.
by
the
from
inches
6
space
given point Q.
joining their
of their
spherical
points
intersectingspheres are
by tangents
AC
A
Def.
all
another
formed
angle
intersection;thus
by
of
radii of two
and
An
plane.
Q.E.D.
the
The
Def.
of what
a
O.
a
circumference
of
sphericalsurfaces
of the two
inches,respectively. The
the
axis MN, generates
revolvingabout
of A
and
1485.
mid-point 0 (" 328).
generates the circumference
A
Find
1484.
Proof
op
at its
AB
the intersection
of
Ex.
_L
MN
locus
P
spherical surfaces is the
two
circumferen
sphericalsurfaces generated by intersecting
0 and
Q revolvingabout line MKf as an axis.
the intersection of the two
sphericalsurfaces the
circumference
Ex.
Theorem
intersecting
two
to the
formed
at their
arcs
is
point
A*
\sT\~
DAE.
angle
two
of circles
arcs
"
"
"^-F
an
intersectingarcs
sphere.*
two
a
is sometimes
attached
to
the
g
expression "spherical angle"
in
V.
Proposition
A
918.
Theorem
is measured
spherical angle
of
the
an
arc
the
by
of
arc
a
angle as a pole and
sides of the angle, prolonged if necessary.
great circle having the vertex
the
intercepted by
423
IX
BOOK
P
sphericalZ APB, with CD
which
is interceptedby
Given
is
P
To
and
that Z
prove
APB
1. Draw
radii OP, OC, and
2.
P
draw
OD
each
and
4.
Prove
OC
IIPR,
5.
But
Z
COD
Cor.
I.
919.
_L
.*.
CD;
of Z
PB
APB.
Z
Proof
PA
and
tangent
PT
to PB.
OP.
IIPT, and
OD
and
PA
pole
OB.
tangent to
OC
oc
of
PR
3. Prove
great O whose
a
CD.
oc
Outline
From
sides
of
iZPT,
Z
hence
i.e.
Z
COD
ARB
Z
=
QC
RPT.
CD.
Q.E.D.
spherical angle is equal to the plane
dihedral
angle formed by the planes of the
A
angle of the
sides of the angle.
920.
a
Cor.
point
on
of all the spherical angles about
the surface of a sphere equals four right angles.
The
II.
sum
of the
corresponding
in plane geometry, frame
definitions of the followingclasses
terms
exact
of spherical
mentary,
angles : acute, right,obtuse, adjacent,complementary, suppleEx.
1486.
By comparison
with
the
definitions
vertical.
Ex.
1487.
Any
Ex.
1488.
If
the
great-circle,
vertical
two
one
circles
sphericalangles are
great circle
are
passes
perpendicularto
through
each
equal.
the
other.
pole
of
another
424
SOLID
AND
LINES
921.
922.
have
PLANES
A
Def.
if,however
one
point.
far
Def.
internally if
neither
one
line
extended, it
spheres
or
the
within
To
to
drawn
to the
plane AB
proofis left as
prove
The
924.
Question.
to make
925.
tangent
and
only
if
other
they
to
a
sphere
is
perpendicular
to
point of tangency.
at
T, and
or
a
radius
OT.
an
exercise for the student.
changes
are
necessary
in the
proof
of
"
313
923 ?
dicular
(Converse of Prop. VI). A plane perpenradius
of a sphere at its outer extremity is
I.
Cor.
to
A.
What
proof of "
it the
each
one
sphere
a
Theorem
VI.
plane AB tangent to sphere 0
the point of tangency.
Given
drawn
to
to
the other.
plane tangent
radius
spherein
tangent
are
Proposition
A
is tangent
plane
the
meets
SPHERE
only one
point in common.
They are tangent
sphere lies within the other, and externally if
sphere lies
923.
a
A
TO
TANGENT
straight
Two
and
one
GEOMETRY
a
to the
sphere.
'
Hint.
See
Ex.
1489.
radius
drawn
Ex.
1490.
"
314.
A
straightline tangent
to the
to a
sphere
is
perpendicularto
point of tangency.
State and prove the
converse
of Ex.
1489,
the
426
GEOMETRY
SOLID
Proposition
inscribe
To
930.
VII.
sphere
a
in
Problem
given tetrahedron.
a
V
tetrahedron
Given
To
inscribe
V-ABC.
sphere in
a
tetrahedron
V-ABC.
I. Construction
A
1. Construct
planesBABS,
whose
are
2.
edges
AB, BC, and
CA,
TCAB
respectively."
691.
0, the
The
will be inscribed
in tetrahedron
II.
V-ABC.
Proof
Reasons
Argument
1. Plane
Z
bisectingdihedral
point of intersection of the three planes,
OF 1. plane ABC.
" 637.
sphere constructed with 0 as center and OF as radius
From
construct
3.
SBCT, and
the
RABS,
AB,
bisector
lies between
edge
By
cons.
and
planes ABV
i.e. it intersects
ABC;
of dihedral
VC
point as D.
intersects plane BCV
plane RABS
line BD
in line AD.
and plane ACV
in
some
2.
.-.
3. Plane
and ABC;
in
a
BD,
4.
lies between
SBCT
line
as
i.e. it intersects
through
planes BCV
plane RASB
between
BA
" 616.
By
cons.
and
BS.
Similarlyplane
RABS
B
in
in
a
TCAR
line
intersects
through
A
plane
between
By steps similar to 1-3.
BOOK
427
IX
Reasons
Argument
and
AB
AD,
intersects
through
5.
6.
But
.-.
.-.
"S
of A
ABB.
SBCT
line
a
A
the
through
pass
BS, and
AR,
in
TCAR
intersect
BS
plane
CT.
as
0, within
as
7.
G
and
AR
plane
and
AR
; and
AR
as
in
terior 5.
in-
Args. 3
"
4.
6.
" 194.
7.
" 617, I.
8.
" 639.
OH.
9.
" 688.
OK.
10.
" 688.
OL.
11.
" 688.
12.
"54,1.
13.
" 925.
14.
"" 926, 927.
point
some
ABB.
CT
in
concurrent
are
point 0.
8.
From
0
draw
OH, OK, and
VBC, and
VAB,
9.
v
0
is in
10.
v
0
is in
11.
v
0
is in
12.
.-.
OF
13.
.-.
each
plane OAB,
plane OB c,
plane OCA,
OH
=
FCM,
OK
=
of the
.-.
sphere
0F
=
OF=
OF
=
OL.
=
faces
four
of the tetrahedron
sphere
0.
in tetrahedron
is inscribed
O
planes
respectively.
is tangent to
14.
_L
OL
V-ABC.
Q.E.D;
III.
The
is left
discussion
Ex.
in
meet
exercise
an
for the student.
dron
planes bisectingthe dihedral angles of a tetraheis equidistant from
the four faces of the
point which
six
The
1501.
as
Discussion
a
tetrahedron.
with
Compare
Hint.
Inscribe
Ex.
1502
Ex.
1503.
.
and
one
Hint.
See
""
258.
sphere
a
volume
The
its surface
"
in
of any
third the radius
491
and
a
given
cube.
tetrahedron
of the
is
inscribed
equal
to the
product
sphere.
492.
point within a triangularpyramid such
planes determined
by the lines joining this point to the vertices
the pyramid into four equivalentparts.
Ex.
1504.
Find
'rfiNT.Compare
a
with
of
Ex.
1094.
that
the
vide
shall di-
428
SOLID
GEOMETRY
Proposition
931.
To
circumscribe
a
VIII.
Problem
sphere
about
a
dron.
given tetrahe-
V
tetrahedron
Given
To
circumscribe
V-ABC.
sphere about
a
tetrahedron
V-ABC.
I. Construction
1.
Through D, the mid-pointof AB, construct plane DO A. AB ;
and
through E, the mid-point of BC, construct
plane EOA.BC;
through F, the mid-point of VB, construct plane FO J_ VB.
2. Join O, the point of intersection of the three planes,to A.
3. The sphere constructed
with 0 as center and OA as radius
will be circumscribed
about
II.
1.
Prove
other
2.
of
V-ABC.
Proof
and
planes OB, OE,
OF
intersect each
lines.
that
Prove
point,as
Outline
that the three
in three
tetrahedron
these
three
lines of
intersection
in
meet
a
O.
3.
Prove
4.
.*.
that
sphere
OA
0
OB
=
=
OC
OV.
=
is circumscribed
about
tetrahedron
V-ABC.
Q.E.D.
932.
a
Are
Questions.
those used
edges of
In
points not in
the
plane
same
determine
sphere.
933.
to
Four
Cor.
the
in
""
dihedral
" 931 could
the
321
and
the
methods
323 ?
In
angles bisected
three edges bisected
used
"
930
be three
be three
in
931
"" 930 and
could
lines
lines
the lines
meeting
lyingin
in
similar
forming
one
the
same
vertex
the
?
face ?
BOOK
at their
in
a
is
point which
a
tetrahedron
the four
equidistantfrom
tices
ver-
tetrahedron.
The
1506.
Ex.
planes perpendicularto the edges of
six
mid-points meet
of the
and
The
1505.
Ex.
429
IX
erected
at their
four vertices of the
four
centers, meet
1507.
Circumscribe
Ex.
1508.
Circumscribe
a
in
point which
a
is
a
tetrahedron,
equidistantfrom
the
tetrahedron.
Ex.
lelopiped. Can
the faces of
perpendicular to
lines
sphere
a
sphere
given cube.
a
about
given rectangular paralrectangularparallelopiped?
a
in any
inscribed
be
about
sphere
a
Explain.
in the
Find
1509.
Ex.
point equidistantfrom
a
SPHERICAL
line
A
Def.
if it
closed
points in space
not
all
is said
to
be
plane.
same
934.
four
POLYGONS
the
on
surface
of
a
sphere
separates a portion of the surface
from
the remaining
of
sphere
portion.
935.
A
Def.
closed
figurecomposed of
its bounding line or
936.
Defs.
surface
more
The
of
arcs
a
A
and
a
of the
is
a
sphere and
lines.
is
boundary
as
circles,
arcs
intersection
of the
ari'gtes
the surface
spherical polygon
of great
the
portionof
sphere whose
bounding
pointsof
a
surface
the
figure on
are
composed
arcs
the sides
are
sphericalangles formed
polygon.
is
of three
the
or
ABCD.
called
of the
figure on
closed
a
of the
the vertices
by
the
polygon,the
of the polygon,
sides
are
the
430
SOLID
937.
A
Def.
diagonal
great circle joiningany
938.
GEOMETRY'
of
spherical polygon
a
an
of
arc
a
non-adjacentvertices.
two
A spherical triangle is
Def.
is
sphericalpolygon"having
a
three sides.
Ex.
1510.
in
terms
By comparison
plane geometry,
of spherical triangles:
and
with
frame
the
exact
definitions of the
definitions of
the
corresponding
following classes
scalene,isosceles,
equilateral,
acute, right,obtuse,
equiangular.
Ex.
With
1511.
a
given
arc
cons.,
step by step, with
as
side, construct
one
an
equilateral
sphericaltriangle.
Hint.
Ex.
Compare
the
With
1512.
three
given
arcs
as
" 124.
sides,construct
a
scalene
cal
spheri-
triangle.
Since
939.
each
side of
great circle (" 936),the
the
of
sphere and
polyhedralZ
form
sphericalpolygon is an arc of a
planes of these arcs meet at the center
at that point a polyhedral angle,as
a
O-ABCD.
This
polyhedralangle and the spherical
polygon are very closelyrelated.
The
of the more
followingare some
student
important relations; the
should
the correctness
prove
940.
polygon
(a) The
sides
have
same
the
of
of each
spherical
a
measures
as
corresponding face angles of
angle.
(b) The anglesof
the
as
Z
the
the
hedral
poly-
sphericalpolygon
a
have
the
same
measures
corresponding dihedral anglesof the polyhedral angle.
Thus,
the
:
same
O-ABCD
sides
AB,
measures
;
and
BC,
as
etc., of sphericalpolygon
face
AAOB,
sphericalA ABC,
the dihedral
A
whose
BOC,
BCD,
ABCD
have
etc., of polyhedral
etc., have
the
same
edges are OB, OC, etc.
These
relations make
it possibleto establish certain properties
of spherical polygons from
the corresponding known
propertiesof the polyhedral
angle,as in "" 941 and 942.
measures
as
BOOK
IX.
Proposition
941.
is
The
the
than
greater
A
spherical
Given
To
of
sum
+
AB
prove
BC
431
IX
sides
two
any
Theorem
of
a
spherical triangle
side.
third
ABC.
"
CA.
Argument
1. Z
AOB
+
Z
2.
Z
^05
oc
^Z?, Z
3.
.-.
^i?
+
BOC
"C"
"
Z
Reasons
COA.
Z
BOCKBC,
CM.
The
less than
Given
To
"
of the
sum
QC
(3.
Q.E.D.
Proposition
942.
CO^
X.
sides
1.
"710.
2.
" 940,
3.
" 362, b.
Theorem
of
any
sphericalpolygon
with
n
360".
sphericalpolygon ABC
prove
'Hint.See
AB
""
+
BC
712 and
+
"""
940, a.
"
a.
"""
360".
sides.
is
432
SOLID
Ex.
In
1513.
80".
Between
Ex.
remaining
Ex.
point
from
C and
If
Ex.
In
AB
is less than
sphericalpolygon
is the
of the
perpendicular bisector
sphere and
in
not
arc
of
is
AB
between
sphericalquadrilateral,
a
lie if three
=
the
sum
of the
arc
CD,
every
unequally
distant
sides
60", 70", and
are
limits must
what
80"?
if three
sides
are
the
40",
70" ?
1517.
Hint.
943.
a
BC
arc
lie ?
CA
arc
40" and
"
D.
side
50",and
side of
arc
surface
1516.
fourth
must
AB
arc
sides.
the
Ex.
limits
Any
1515.
on
sphericaltriangle ABC,
what
1514.
GEOMETRY
side of
Any
See Ex.
a
sphericalpolygon
is less than
180".
1514.
angle
If,with A, B, and C the vertices of any sphericaltrias
poles,three great circles are constructed,as B'c'ED,
C'A'd,and EA'b',the surface of the sphere will be divided into
the hemifour of which
sphere
are
on
seen
eight sphericaltriangles,
angles,
representedin the figure. Of these eight sphericaltri^'.B'C'is the
and
A1 lie
AC, and
C
the
on
C'
and
the
that is
only one
side of BC,
same
on
and
one
B
side of
same
B'
and
Questions.
is entirelyoutside
(a)
A'B'C
is
(")
A'B'C
is
Ex.
whose
are
figureabove,
Can
ABC.
entirelywithin
partly outside
What
1518.
sides
of A
In the
is the
quadrants
?
the
ABC
of and
two
A
same
A
side of
angle
particulartritriangleABC.
A'B'C,
A
the
that
This
AB.
a'b'c' is called the polar triangle of
944.
on
situated
so
be
so
the
polar of
constructed
A ABC,
that
:
?
partly within
polar triangleof
a
ABC?
sphericaltriangleall
of
434
SOLID
GEOMETRY
Proposition
In
947.
side of the
that
Theorem
triangles each
polar
two
XII.
other
of
which
its vertex
is the
and
one
pole
are
to 180"-
together equal, numerically,
E
angle of
a
with sides denoted
polar A ABC and A'b'c',
by a, b, c,
a',b',c',respectively.
(a) Z ^4-a' 180", Z B+b'
l$0", Z c+c'= 180";
prove:
Z C'+c=180".
(6) ZA'+a=180", ZB'+b=180",
Given
and
To
=
(a)
1.
Argument
Let
B'C' at
and
D
E,
4.
C'E +
"
.
But
5.
6.
Z
5
+ a'
=
+ "'
=
(6) The proof of (6)is
Hint.
948.
Let
BC
prolonged
Question.
In
frequentlyspoken
Ex.
Find
Ex.
1520.
The
the sides of its
1521.
equiangular ;
If
and
=
=
180".
A.
Z
180", and
meet
as
of
C
+ c'
exercise
an
A'B'
history
as
a'
+
180".
left
the
of
180" ; i.e. ED
=
of Z
measure
arc
180".
=
+ ZXB'
EB
.\Z^
Likewise
c'D + EB'
+
ED
is the
ED
AC
=
.-.
.
Only
(prolongedif necessary) intersect
90".
90" and EB'
respectively
; then c'D
and
AB
arcs
2.
3.
=
at
.ffand
180".
q.e.d.
for the student.
A'C
mathematics,
supplemental
=
at
why
if.
are
polar
angles
tri-
triangles ?
angles of a sphericaltriangleare
polar triangle.
75",85", and
145".
is
its polartriangle
sphericaltriangleis equilateral,
conversely.
a
BOOK
XIII.
Proposition
The
949.
180"
greater than
sphericalA
Given
To
ZA
prove
tiie
of
sum
Theorem
angles of
and
less than
ABC
with
+ Zb
435
IX
+ Zc"
spherical triangle is
540".
a
sides denoted
180" and
by
b, and
Reasons
by a',
1.
" 943.
b',and c',be the polar of A ABC.
Z A + a'
2. Then
180", Z B +V
180",
2.
" 947.
3.
" 54, 2.
c' "360".
4.
" 942.
"7"180".
5.
" 54, 6.
6.
" 938.
7.
" 54, 6.
1. Let
A
sides
denoted
'
=
.:ZA
4.
But
5.
.:ZA
6.
Again,
7.
.-.
or
obtuse
951.
has
ZB+Zc
a' +
+
b' +
+ ZB+Z
Z^
950.
two,
+
=
180".
Zc+c'=
3.
+
c.
540".
"
Argument
A'b'c',with
a,
(a'+b'+c')
=
a1 +
b' +
c' "
0".
Z^B
+ Z
c"
540".
54:0o.
q.e.d.
be one,
spherical triangle there can
three right angles ; there cajv be one, two, or three
angles.
Cor.
In
a
Throughout the Solid Geometry
Note.
student's
attention
definitions and theorems
called to the relations between
constantlybeen
the
of plane
corresponding definitions and theorems
In the remaining portion of the geometry of the sphere there
geometry.
be
of these comparisons, but here the student
will likewise be many
must
of solid
geometry
and
the
his guard
on
particularly
for
contrasts
as
well
as
comparisons.
For
ex-
436
SOLID
while
ample,
180", he
the
learned
has
be
may
have
but
two,
or
any
rightor
one
three
and
952.
circles
a
953.
Def.
angles is
Thus
Ex.
great circles perpendicular to
the
equator, it will
definite.
sphericaltrianglesmore
rightangles
or
1522.
its angles right
sphericaltriangle
having all of
trirectangular spherical triangle.
a
meridians,as NA
obtuse Z, form
an
If the
becomes
ANB
angle, a sphericaltrianglemay have one,
angles (" 950).
one, two, or three obtuse
that, considering the earth as a sphere, the
earth
the poles of the equator, and that
are
trianglecontainingtwo
spherical
A
called
two
acute
to
birectangular spherical triangle.
sphericalA.
A
of the
are
A
Def.
is called
an
poles
sum
180"
plane triangleis equal to exactly
of the angles of a sphericaltriangle
540" ; while a plane trianglecan
obtuse
will recall
thinkingabout
his
make
one
a
the
that
from
rightangles or
south
all meridian
angles of
(" 949)
number
If the student
north
of the
sum
GEOMETRY
a
Z
and
with
between
making
NB,
the
equator
and
NA
at the north
NB
a
is
pole
birectangular
made
rt. Z,
a
sphericalA.
trirectangular
kind
What
of
arcs
ANB
spherical angle
spherical triangle quadrants?
measures
are
?
Are
What
NA
and
two
NB
Then
?
sides of any
is each
side
of
a
what
arc
birectangular
trirectangular
sphericaltriangle?
Ex.
1523.
triangleis
Ex.
If two
sides
of
sphericaltriangleare
birectangular. (Hint. Apply " 947.)
1524.
What
is the
a
polar triangleof
a
quadrants, the
trirectangularspherical
triangle?
Ex.
sum
Make
1525.
of the two
this
new
trianglewhose
exterior
angle of
sphericaltriangleis less than the
interior angles. Compare this exercise with " 215.
remote
fact clear by applying it to a birectangular spherical
third angle is : (a) acute; (6) right; (c) obtuse.
An
a
BOOK
XIV.
Proposition
Theorem
in the
In
equal spheres,or
triangles are equal :
954.
437
IX
sphere,two spherical
same
I.
the two
If a side and
adjacent angles of one are
equal respectivelyto a side and the two adjacent angles
of the other;
II.
If
sides
two
the
and
equal respectivelyto
included
sides
two
angle
of
one
tlie included
and
are
angle
of the other;
III.
the
If
the
provided
each
In
left
II,and
and
thus
in
"
954 ?
956.
""
105
and
107.
Is the method
Two
Def.
:
cases
here
order.
same
for the student.
corresponding trihedral
sphericalA are equal.
the
prove
that the
Prop. XIV,
Could
in the
arranged
are
show
Compare
equal respectivelyto
are
one
exercises
above
Questions.
with
as
of the
equal ("" 702, 704) ;
955.
of
equal parts
proofs are
Hint.
sides
sides of the other
the three
The
three
I and
the methods
II, with
used
there
suggested preferable? why
spherical polygons
correspondingpolyhedralangles are
" 702,
be
I and
employed
?
symmetrical
are
A
if the
symmetrical.
^"
The
of
following are some
triangles;the student
spherical
957.
of each
the
should
equal,but
arranged
(b) Two
isosceles
Prove
trihedral A
prove
the
ness
correct-
:
(a) Symmetrical sphericaltriangleshave
Hint.
of symmetproperties
rical
in
reverse
their parts
respectively
order.
are
symmetrical sphericaltriangles
(6) by superpositionor by showing
are
equal.
that the
equal.
ing
correspond-
438
SOLID
GEOMETRY
Proposition
958.
In
XV.
Theorem
equal spheres, or in the same
sphere, two
spherical triangles are equivalent.
A
spheres O
To
A'
symmetrical sphericalA
Given
ABC
A'B'c'
and
sphericalA
A
spherical
ABC=^
2.
3.
and P' be
Reasons
polesof
small "
through
A, B, C, and A',B',C',respectively.
Arcs AB, BC, CA are equal,respectively,
to arcs
A'B',B'C',C'A'.
.-.
P
chords
AB, BC, CA,
to chords
4.
.-.
plane
5.
.-.
O
6.
Draw
equal
A'b'C*.
Argument
1. Let
in
O'.
and
prove
metrical
sym-
A
ABC
ABC
equal,respectively,
plane A
=
of
908, h.
2. "
957, a.
3. "
298, II.
A'b',b'c',c'a'.
a'b'c*.
4.
"116.
5. " 324.
O A'B'C'.
=
arcs
are
1. "
great (D PA, PB, PC, p'a',
6. "
908, g.
p'b',and P'c'.
.
7.
PA
8.
.-.
=
9.
11.
PB
A
spherical
symmetrical.
.'.
Likewise
.-.
P'A'=
PC=
=
isosceles
are
10.
7. " 913.
Then
A
APB
A
BPC
A
CPA
A
spherical
p'b'=
APB
and
==
A
A' P'B'.
=
A
b'p'c' and
=
A
C'P'A'.
ABCo
p'c'.
a' p'b'
8. "956.
9. "957, b.
10.
lar
By stepssimito 8-9.
Aa'b'c'.
spherical
Q.E.D.
11.
" 54, 2.
Proposition
439
IX
BOOK
XVI.
Theorem
sphere, two spherical
equal spheres,or in the same
and
thereforeequivalent :
triangles are symmetrical,
In
959.
I.
If
the
and
side
a
a
side and
and
the
equal respectivelyto
of the other ;
II.
//
sides
two
adjacent angles of
two
the two
one
are
adjacent angles
included
angle of
one
are
the included
angle
equal respectivelyto two sides and
of the other;
III. If the
three
sides of one
are
equal respectively
to the three sides of the other :
provided
the
equal parts
left
proofsare
The
In
Hint.
of the
each
as
cases
thus
show
symmetrical (" 709) ; and
bisector
The
1526.
"Ex.
exercises
above
of
that
Ex.
The
1527.
triangleto the
arc
drawn
mid-point
from
corresponding trihedral A
spherical" are symmetrical.
the
the
angle
sphericaltriangleis perpendicular to
order.
reverse
for the student.
prove
the
in
arranged
are
of
at
the
vertex
the
base
and
bisects it.
the
vertex
of the
base
bisects
prove
the
theorems
the
plane
of
the
an
isosceles
an
angle
vertex
isosceles
spherical
and
is perpendicular
to the base.
Ex.
to
State
1528.
the
following theorems
(1) Every point
from
the
and
in the
on
:
(" 134).
(2) Every point equidistantfrom the ends of
bisector of that line (" 139).
(3) Every point in the bisector
sides of the angle (" 253).
PD
"
In
AB
Ex.
lay off BE
The
perpendicular to
that
Ex.
each
are
triangles
; and
In
=
line is
equidistant
of
an
angle
a
line lies in the perpendicular
is
equidistant from
the
BD.
diagonals of
corresponds
1530.
a
figurefor (3),corresponding to the figureof " 252, draw
the
and
1529.
of
perpendicular bisector
of that line
ends
Hint.
ing
sphere correspond-
the
on
other.
an
Prove.
equilateralsphericalquadrilateralare
State the theorem
in
plane geometry
to this exercise.
equal spheres,
in the
sphere, if two spherical
their polar trianglesare mutually equimutually equilateral,
angular
conversely.
or
same
440
GEOMETRY
SOLID
Proposition
In
960.
XVII.
in
equal spheres, or
Theorem
the
same
sphere, if
two
spherical triangles are mutually equiangular, they are
mutually equilateral, and are either equal or rical.
symmet-
sphericalA T and T* in equalspheres,or
sphere,and mutually equiangular..
Given
To
Tf mutually equilateral
and
and
T
prove
in the
either
same
equal or
symmetrical.
Reasons
Argument
Let
A
and
T
and
Then
.*.
and
P
'
"943.
polars of A T
T',respectively.
T' are mutually equiangular.
By hyp.
and
P'
P
are
lateral. "947.
mutually equiP
and
P
P'
be
the
either
are
hence
and
equal or symmetrical,4. "" 954, III,
and 959,III.
are
mutually
equiangular.
T' are mutually equilateral.
.*. T and
.-.
T
and
T'
either
are
equal
or
metrical.
symQ.E.D.
Ex.
1531.
what
Ex.
1532.
from
points in
two
space,
plane geometry,
In
can
be said of them
the
Find
on
Are
Same
as
reason
4.
angular,
trianglesare mutually equithey equal ? equivalent?
locus of all points of
a
the
surface
of the
the surface
of the
sphere.
given points on
not
?
if two
"947.
tant
sphere that are equidisfrom
two
given
sphere ;
442
SOLID
GEOMETRY
Proposition
963.
If
sphericalA
Given
BC"
prove
Theorem
spherical triangle are
a
greater
angle
is
Z
C.
Draw
an
2.
witli Z
ABC
A
"
3. But
4.
.
of
great O
a
1. " 908, g.
making
AD
BD
+
'dc.
2.
" 962.
a1"" AB.
3.
" 941.
4.
" 309.
=
+
bd
\
Reasons
a
Td
Then
the
AB.
arc
Z1=Z
than
greater
Argument
1.
unequal,
less angle.
side opposite the
To
the
opposite
side
the
angles of
two
XX.
DC
; i.e. BO
AB
"
"
AB.
Q.E.D.
birectangularsphericaltrianglethe side included by
the two
right angles is less than, equal to, or greater than, either of the
other two
sides,according as the angle oppositeis less than, equal to, or
In
Ex.
1533.
greater
than
Ex.
1534.
An
Ex.
1535.
If
dihedral
a
90".
equilateralsphericaltriangleis also equiangular.
two
face
angles oppositeare
angles
equal.
of
a
trihedral
angle
Ex.
1536.
State and
prove
the
converse
of Ex.
1534.
Ex.
1537.
State and
prove
the
converse
of Ex.
1535.
Ex.
1538.
The
triangleform
Ex.
inches
solid
Ex.
The
1539.
and
an
the
bisectingthe base angles of
isosceles sphericaltriangle.
arcs
of
bases
altitude 3
isosceles
an
inches
; find
the
total
generated by revolvingthe trapezoidabout
1540.
Find
the
total
revolvingthe trapezoid of
Ex.
area
1530
and
about
volume
its shorter
solid
base
and
as
as
6
of the
volume
longerbase
of the
cal
spheri-
inches
14
and
area
its
isosceles
an
trapezoid are
equal, the
are
an
axis.
generated by
an
axis.
XXI.
Proposition
964.
If
of
sides
two
443
IX
BOOK
Theorem
spherical triangle
a
angle opposite the greater
angle opposite the less side.
the
side
is
are
greater
unequal,
than
the
B
sphericalA
Given
To
Z
prove
The
A
proof is
Hint.
Prove
965.
"
left
in
Ex.
1525.
Ex.
156
"
?
Does
If
1541.
C.
as
an
between
the
greater sides.
Hint.
Compare
Ex.
as
Ex.
rt. A
VAO
VA
:
=
total
sphericalA
?
See
spherical quadrilateralare
a
1539
Ex.
volume
and
area
about
Ex.
1544.
VBC
VO,
and
VA,
then
Pind
of
the
the
of the
solid
generated by
of its
perpendicular bisector
VB
scribed
sphere in-
edge
is
a.
sphere, A
the
the
of face EDF.
center
that
sphere. Show
similar.
are
tejtrahedron whose
In the
B
inscribed
of
VC,
whose
center
and
VED,
radius
VB.
Find
the
of the
radius
the
0 be
and
1328).
Hin*.
of
in
153.
regular tetrahedron
Let
OA
=
a
of face
Then
proof hold
that
proved by the method
other two
Ex.
the
Pind
1543.
Hint.
center
been
have
axis.
an
in
XXI
sides,the sphericalangle included
sides is greater than the sphericalangle between
revolvingthe trapezoid of
bases
4 of
the
with
Find
1542.
for the student.
adjacent sides
two
shorter
two
Prop.
reason
two
AB.
"
indirect method.
than
greater, respectively,
the
BC
exercise
Could
Questions.
used
Z
the
by
with
ABC
Then
can
be
VO
found
:
VC
(Ex.
OA.
the
radius
is
of the
edge
figureof Ex. 1543, draw
sphere circumscribed
a.
AD
and
OD.
about
a
lar
regu-
444
SOLID
GEOMETRY
MENSURATION
OF
SPHERE
THE
Areas
Proposition
966.
If
XXII.
Theorem
isosceles
triangle is revolved about a straight
line lying in its plane and
passing through its vertex but
not intersecting its surface, the area
of the surf ace generated
the
the
is
to
base
the product
of
equal
by
triangle
of
its
an
the
projection on
circle whose
axis
is the
radius
the
and
altitude
x
circumference
of
a
of the triangleX
X
c
OH
D
Y
Y
Fig.
line
not
of
lying in the plane
intersectingthe surface
by
AB
2.
base
AB
of
A
of A
AB
area
is not
=
CD
"
E
2.
the
Since
frustum
of
a
OE,
passing through
AOB,
and
surface
a
O
3.
str.
and.
the
projection
generated by AB be
CD
2 irOE.
and
IIXY
does
Argument
1. From
altitude
and
AOB
of the
area
Fig.
AB.
area
prove
I. If
let the
J7;
on
denoted
To
with
AOB
XY
AB
Fig.
1.
isosceles A
Given
Y
draw
EH
A.
surface
rt. circular
3.
From
A
4.
Then
in rt. A
draw
not
meet
(Fig.1).
XY
Only
XY.
generated by
cone,
AK"
BD.
BAK
and
area
OEH, Z
AB
is the
AB
=
BAR
AB
=
"
Z
surface
2 -n-EH.
OEH.
of
a
BOOK
5.
.-.
ABAK
6.
.*.
AB.AK"
7.
But
8.
.-.
II.
CD;
AB
proofs of
Hint.
967.
Cor.
CD
-EH=
"
"
OE.
OE.
and
q.e.d.
lies in
pointA
III
left
are
proof given
If half of
as
for I will
(Fig.2).
XY
of sides is circumscribed
area
of the surface generated by
the
diameter
revolves
about
of
as
exercises
apply
to
for the student.
Figs.2
regular polygon
a
number
the semicircle
AK
=
-2 -nOE.
II and
I.
EH
"
(Fig.8).
if the
See
AB
.'.
IIXY
IIXT
^B
i.e. AB
EH)
CD
"
is not
AB
III. If
The
OE:
area
If
AOEH.
~
AK=
445
IX
about
its
a
and
3.
with
an
even
semicircle, the
semiperimeter
it
as
axis, is equal
an
product of the diameter
the
of the regular polygon and
circumference of a circle whose
the
to
radius
is
the
R,
radius
of
the
given semicircle.
Outline
of
Proof
1. Area
A'b'
=
A'f'
area
B'c'
=
f'o'
2.
.-.
968.
number
area
a'b'c'--
Cor.
n.
of
-2ttR;
2 ttR; etc.
"
(a'f'+
=
If half of
sides
f'o' +
is inscribed
in
a
surface generated by its semiit revolves
about
perimeter as
diameter
of the semicircle
axis, is equal
an
the
to the
as
product of
diameter
of the semicircle
and
the circumference of a circle
whose
radius
is the apothem
of
the regular polygon.
'
ifint.
Prove
area
ABC
"
"
"
=
AE
ttR
=
regular polygon
a
the
the
)2
"
2
ira.
A'E'
with
semicircle, the
"
2 ttR.
an
area
even
of
446
SOLID
969.
same
Cor.
If halves of regular polygons with the
of sides are circumscribed
about, and
m.
number
even
inscribed
GEOMETRY
in, a semicircle, then
by repeatedly doubling the
number
of sides of these polygons, and making the polygons
always regular, the surfaces generated by the semiabout
the
perimeters of the polygons as they revolve
diameter
axis approach a comof the semicircle
as
an
mon
limit.
Outline
1.
the
If
and
S
surfaces
Proof
of
denote
s
the
generated by the semi-
perimeters A'b'c'--- and
they revolve about a'e'
then
6'
=
A'E'
and
s
=
AE
2.
2 *-#
"
2
"
ttcl
A'E1
S
.-.
of
areas
axis,
an
as
as
("967) ;
("968).
A'E'
2 ttR
"
ABC---
=
-
AE
8
But
2
"
AE
7ra
polygon a'b'c'-A'E'
polygon ABC--
"
(" 438).
a'b'
-(""419,435).
V
=
AE
AB
a
s
a2
a
R2
S
Rz
S
7.
.-.
by steps
970.
Which
Def.
the
Args.
to
6-11
(" 855), S
and
limit.
common
a
similar
The
surface
successive
s
proach
ap-
q.e.d.
of
sphere
a
surfaces
is the
limit
common
generated by halves
of
regular
number
of sides approach,if these
even
polygonswith the same
semipolygonsfulfill the followingconditions :
about,and inscribed in,a
(1) They must be circumscribed
great semicircle
(2) The
each side
of the
number
of
sphere;
sides
approachingzero
as
must
a
be
limit.
successivelyincreased,
XXIII.
Proposition
The
971.
Theorem
of the surface of a sphere is equal to four
of a great circle of the spliere.
area
the
times
447
IX
BOOK
area
A
sphere
Given
of its surface
To
denoted
S
prove
with
0
denoted
its radius
by
by R,
S.
4 -irR2.
=
Reasons
Argument
1.
semicircle
the
In
half
of
a
inscribe
ACE
ABCDE,
with
regular polygon
number
even
the
and
sides.
of
Denote
1. " 517,
a.
an
its
of the
apothem by a, and the area
surface generated by the semiperimeter
it revolves
as
about
AE
as
an
axis
bys'.
S'
=
AE
i.e. S'
=
2 R
2. Then
3.
As
the
number
polygon, of
2
"
"
2
2.
ira
;
ttcl
of
=
4 irRa.
sides
which
as
a
limit.
4.
Also
a
5.
.-.
approachesR as
a
approaches4
a
as
6.
But
1." V- 5
"
is
S'
a
regular
"970.
half,is
approaches S
" 543, I.
limit.
irR
"
R, i.e. 4 irR\
" 590.
limit.
S' is
=
of the
ABCDE
repeatedlydoubled,
4 nR
" 968.
always equal to
4 irR2.
4 7ri?
"
Arg.
a.
Q.E.D.
2.
" 355.
area
448
GEOMETRY
SOLID
972.
Cor.
to each
The
I.
other
the
as
Outline
1.
S
47ri?2andS'
=
R2
.-.?-i^47ri2'2;
=
=
Historical
"
=
8'
"
'"'
".
=
R12
4tt2*'2
Prop. XXIII
Sphere and
=
-
"
S'~D'2'
D'2'
(2rJ~
Note.
entitled
treatise
the
as
D*
(2 R^
=
=
R,2~ "R"2~
973.
the
4 R2
But
U
'
and
are
Proof
of
'
2
spheres
diameters.
their
of
squares
of the surfaces of two
of their radii
squares
areas
is
given
in
XXXV
Prop.
as
Cylinder by Archimedes, already spoken
of in S 809.
is
Ex.
1545.
Find
Ex.
1546.
What
1| decimeters, at
Ex.
from
the
of
surface
will it cost
an
The
1547.
inches
the
of
area
is 3000
center
of
" of
of
section
a
ir
whose
a
a
cent
radius
globe whose
a
centimeter
square
per
sphere made
inches.
square
is 10 inches.
diameter
gildthe surface of
to
cost
average
sphere
a
Find
by
11
plane
a
of the
surface
the
?
sphere.
whose
edge
whose
surface
the
radius
is twice
the
Find
total surface
of
each
are
The
1550.
Ex.
Ex.
equal
about
a
cube
of the
a
of
given sphere
sphere
a
of
the radius
Find
one
;
half ;
diameter
whose
right circular cylinder whose
altitude
nth.
one
is 2
and
sphere
a
R, and
diameter
results
the
of
Ex.
of
surface
a
1550
state, in the
sphere
to the
of
form
total surface
a
of the
cylinder.
Show
1552.
to
1550, the surface
the lateral surface of the cylinder.
that,
Historical
the surface
that
sphere is R.
a
surface
the relation of the
exactly equal
974.
of
the surface
From
circumscribed
Ex.
sphere circumscribed
a
2 R.
to
1551.
theorem,
of
is 12 inches.
1549.
Ex.
the surface
Find
1548.
Ex.
of
in
Note.
a
Ex.
discovery of
The
sphere is
thirds
two
the
of the
of the
sphere is
remarkable
surface
property
of the
scribed
circum-
The
1551) is due again to Archimedes.
discovery of this proposition,and the discovery of the corresponding
the philosopher'schief pride, and
(" 1001), were
proposition for volumes
cylinder (Exs.
he
therefore
asked
wishes
tomb.
His
further
account
that
were
1550
a
and
figure of
carried
of Archimedes,
this
out
read
by
also
proposition be inscribed
his
friend
" 542.)
Marcellus.
on
his
(For
a
450
SOLID
980.
the
Cor.
zones
Question.
In
of two
Ex.
equal spheres, or
of two
areas
981.
In
III.
lines.
The
1553.
Use
" 980
of
is the
979
""
1554.
Show
that
Ex.
1555.
Find
the
between
The
1556.
Ex.
planes divide
of the
four
Ex.
base
other
is
the
as
equal
to the
of
of the
area
and
diameter
of
surface
radius
the
971 is
"
a
of
of the
specialcase
a
if the
zone
sphere
sphere is 16 inches.
four equal parts.
Find
into
of
Prove
that
distance
is 6 inches.
Three
a
the
half of the earth's surface
one
Considering
1558.
one
the earth
twice
Is the
Ex.
1560.
of the
sphere with
a
as
parallel
of
area
the
areas
equal ?
the
sphere
north
radius
B, find
bases
observer
Let
must
1562.
radius
B,
60" from
find
the
north
?
as
a
that
the
of
center
so
the
sphere
a
whose
of the
sixth
one
Are
equator.
the
radius
surface
the
45c
two
is B
of the
be
E
=
"
.
the
Find
from
What
the
OB, " then
portion
if the
seen
of
eye
of
the
observer.
use
II.
" 443,
,
w
surface
of
distance
of the eye
of
sphere
is 2B
the
center
of
The
radii
of two
the
a
the
?
nB?
Ex.
1563.
inches.
area
be
can
observer
SB?
the
be
sphere with
north pole ;
a
45" from
pole ; (6) 30" and
Explain your answer.
an
Is
;
"
the
7?
sphere with radius i?,find
parallelsof latitude: (a) 30" and
are
far from
as
the
earth
3
sphere
30" from
Considering the
of
2
is visible ?
AB
Ex.
earth
other
How
eye
Hint.
Then
30"
?
twice
area
whose
1561.
Ex.
must
each
lies within
altitude is
pole,whose
the
one
zone
from
other
extending
zone
pole.
the
Considering the
1559.
of the
north
adjoining the
zone
area
Ex.
area
979.
"
thus formed.
zones
of the
area
a
zone.
7?
the
of
area
equator.
Ex.
area
ucts
prod-
Explain.
generating the
arc
the formula
this diameter
1557.
of the
of the
their altitudes.
as
this rule ?
one
sphere,
same
444, II.
is 8 inches
its bases
of
zone
chord
and
Ex.
a
the
to each
are
exception to
an
in
other
to each
are
general,surfaces
area
radius
circle whose
Hint.
Is
GEOMETRY
A
plane
of the section
of the surface cut
is
concentric
spheres are
6 inches
and
10
passed tangent to the inner sphere. Find : (a) the
of the outer
sphere made by the plane ; (") the area
off of the outer sphere by the plane,
982.
A
Def.
the surface of
figureon
closed
a
a
sphere
composed of two
of great circles,
as
is
boundary
whose
is
lune
451
IX
BOOK
semicircumferences
NASB.
983.
and
NAS
vertices
the
NBS;
of
the
sides
of
the
lune, as
sphericalangles
angles
of the
lune,
pointsof
section
inter-
the
of
are
N
formed
the sides of the
by
ences
semicircumfer-
two
called the sides
are
as
The
Defs.
lune,
A
the
S-, the
and
at the vertices
lune
as
called
called the
are
and
ANB
BSA.
the followingproperty of lunes :
Prove, by superposition,
In equal spheres,or in the same
sphere,two lunes are equal if
984.
their
anglesare
So
985.
equal.
considered
far,the surfaces
in terms
in connection
with
the
of square
units,i.e. square
inches, square feet,etc. For example, if the radius of a sphere
i.e. 144
is 6 inches, the surface of the sphere is 4"7r62,
square
inches.
But, as the sides and angles of a lune and a spherical
polygon are given in degrees and not in linear units,it will be
unit for determining the areas
to introduce
new
some
necessary
of these figures. For this purpose the entire surface of a sphere
is thought of as being divided
into 720 equal parts,and each
of these parts is called a spherical
one
degree. Hence :
sphere have
been
measured
ir
986.
Def.
Now
if the
A spherical degree is yj-gof the surface of
area
of
a
lune
of
or
a
a
sphericaltrianglecan
sphere.
be obtained
in
can
sphericaldegrees,the area
easilybe changed to
of a
units.
that the area
For example, if it is found
square
sphericaltriangleis 80 sphericaldegrees,its area is -f^, i.e. ^
of the entire surface
of the sphere. On
the sphere whose
radius is 6 inches,the area
of
of the given trianglewill be -J-
144
7r
square
theorems
on-
are
tye surface
inches,
i.e. 16
for the purpose
of
a
spherein
inches.
"*
square
of
determining the
terms
of
The
following
of
figures
degrees.
spherical
areas
452
SOLID
GEOMETRY
Proposition
The
987.
tl%e
sphere
of
area
the
as
XXIV.
lune
a
Theorem
is to the
number
of the surface of
area
in
of degrees
angle of
the
is to 360.
lune
N
N
Q
Given
by
the
lune
NASB
area
denoted
Nf its
sphere denoted
with
by
L
E
the number
let O EQ
S ;
the
and
by L,
degreesin
of
of
area
its Z
the surface
great O whose
be the
denoted
of the
pole is
N
m
To
prove
=
-
360
8
I.
If
circumference
and
AB
arc
EQ
commensurable
are
(Fig.1).
Reasons
Argument
1.
Let
be
m
a
common
and
circumference
that
m
is contained
Through
the
1.
AB
circumference
suppose
r
times
" 341.
~t'
pointsof
division
EQ
semicir-
pass
great circles from
of
" 335.
t times.
EQ
EQ
several
cumferences
AB
arc
arc
r
circumference
on
in
AB
arc
Then
and
EQ,
in circumference
and
of
measure
"
908, h.
N
to S.
4.
Then
and
lune
the
NASB
is divided
surface
of the
lunes,each equal to
lune
into
r
sphere
NCSB.
lunes
into
4.
t
"984.
N.
453
II.
If
and
AB
arc
circumference
EQ
incommensurable
are
(Fig.2).
The
proof is
The
Hint.
988.
left
proof
is similar
degrees, is equal
angle.
of
lune, expressed in spherical
a
the
twice
Outline
L
360
8
is 35".
Ex.
and
Find
1564.
Ex.
Find
each
as
other
Ex.
area
the
the
as
In
1568.
a
the
of
area
is
lune
a
zone
sphericaldegrees if its angle
surface of the sphere ?
inches
in square
8 inches.
(Use
in the
2n.
tr
=
if its
-2/.)
sphere, two
same
is 42"
angle
lunes
to
are
unequal spheres,but with equal angles,are
of the radii of their spheres.
squares
a
lune
sphere
angle is
whose
visible
radius
whose
Considering the
1569.
of the
in
=
to
lunes in
Two
equivalentto
Ex.
lune
a
equal spheres,or
their angles.
1567.
Ex.
of
L
360*
720
In
1566.
other
N
**
area
of the sphere
the radius
each
the
Us
in
Proof
of
part is the lune of the entire
What
1565.
Ex.
("987).
J
V
of degrees
number
L
JL
=
of " 409, II.
to that
area
to
for the student.
exercise
an
The
I.
Cor.
as
from
a
is R, find the
altitude
of
a
zone
45".
earth
as
a
point
at
a
sphere with radius B, rind
the surface
height h above
the
of
earth.
989.
Def.
The
thre^excess of the
spherical
sum
of its
excess
of
angles over
a
sphericaltriangleis
180".
454
GEOMETRY
SOLID
Theorem
XXV.
Proposition
of a spherical triangle, expressed
spherical degrees, is equal to its spherical excess.
The
990.
area
in
B
r
i
i
/
A
A/
J
/
-"
:
\
sphericalA
Given
by
/ /A
with
ABC
its
spherical excess
E.
To
of A
area
prove
ABC
sphericaldegrees.
E
=
Argument
1.
Complete the
CA
2.
Aab'c'
3.
.-.
AAB'C'
4.
.-.
A
5.
.-., expressed in
.-.
of which
" 958.
+ A AB'C'o
A
-{-A A'BC.
ABC
ABC
+
A
AB'C
A
ABC
+
A
ABC'
lune
A
=
=
lune
B
=2
=
lune
C
f"=
=
2A^LBC-f-(A^BC+AAB'(7'
-f A ABC')= 2(A + B+ C).
A ABC+
9.
.-.
A
surface
ABC+
^BC
A
of
360
+ 180
^z?c
=
B-,
2 C.
+ A^LB,C
=
360.
hemisphere
2 (A + 5 + c).
=
=
^
+
5
+
a
6')- 180,
sphericaldegrees.'
A
2 A ;
AB'C'+ A AB'C+AABC'
a
(i+B
+
i.e.
" 54, 2.
" 988.
sphericaldegrees,
A
2 A
" 956.
symmetrical.
oAA'BC
AB'C'
.-.
" 908, g.
arcs.
are
+ A
8.
".
A'BC
ABC
=
.
are
A
But
10.
and
ABC
Reasons
circumferences
AB, BC, and
6.
denoted
^
q.e.d.
BOOK
" 949 it
In
991.
proved that
was
sphericaltriangleis greater
a
0" to
from
of
sphericalexcess
the
Hence
360", from
455
IX
the
than
180"
angles
less than
and
(" 990) that
the
of
540".
sphericaltrianglemay
a
it follows
which
of the
sum
vary
of
area
yf^ to ff" of the entire
of a sphericaltrianglemay
surface ; i.e. the area
vary from
nothing to \ the surface of the sphere. Thus in a spherical
the
trianglewhose angles are 70", 80",and 100",respectively,
the
70" ; i.e.
180"
is (70"+ 80" + 100")
sphericalexcess
of the given triangleis -f^ of the surface of the sphere.
area
sphericaltrianglemay
a
from
vary
=
-
992.
wrote
treatise in which
a
is
there
although
no
he describes
propertiesof
The
their solution.
at
attempt
the
Alexandria
(circ.98 a.d.)
sphericaltriangles,
expression
for the
1626
first given about
sphericaltriangle,as stated in " 990, was
also discovered
was
by Girard.
(See also " 946.) This theorem
by Cavalieri,a prominent Italian mathematician.
of
area
a.d.
a
two, into
other
divided
?
Then
sphericaldegrees ?
Ex.
Test
Find
1571.
Ex.
has
an
of 216
area
of square
number
trianglewhose
and
its
1575.
Ex.
and
the
Ex.
a
1577.
area
of
a
surface
inches, find
of
the
the
face
sur-
the
area
of
120".
60", 120", and
sphericaltriangleare
inches.
in the ratio
is 16
diameter
70",86", and
a
part of the
What
sphericaltrianglein a sphere whose
Find
feet are
95", 105",and 130".
of the triangle.
whose
angles of
The
Find
the
radius
sphericaltriangleis
of 2, 3, and
5.
Find
of the
90
the
sphere.
160",
(Use
sphericaldegrees,
angles.
angle (1) of an equilateralspherical triangle,
equivalentto one third the surface of a sphere.
Find
lune, each
area
sphere
lOOf square
angles are
1576.
Ex.
(2) of
area
is
square
angles are
The
1574.
Ex.
a
is 80".
a
feet in the
In
1573.
Ex.
spherical degrees of a birectangular
angles is 70" ; of an equilateralspherical
of whose
angles of
The
XXV.
in
area
angles
triangle?
1572.
are
by applying Prop.
answer
your
the
sphericaltriangleone
of whose
triangleone
sphere is each
dependent
in-
drawn, each perpendicular to the
how
trirectangularsphericaltrianglesis the surface
many
of a trirectangular
what
is the area
sphericaltrianglein
If three great circles
1570.
Ex.
a
of
Menelaus
Note.
Historical
Find
one
spherical
triangle
the
the
of
angle of a lune equivalentto
whose
angles is 84".
an
equilateral
456
SOLID
GEOMETRY
Proposition
993.
The
of
area
spherical degrees, is
by 180"
has
To
spherical polygon, expressed in
ished
equal to the sum
of its angles dimin-
taken
as
times
many
angles by
of
area
prove
degrees,
(n
T"
=
less two
"
ABCD
"
"
"
with
n
gon
poly-
the
T.
polygon
2)180.
ABCD
"
"",
expressed in spherical
Reasons
such
vertex
any
the
sides ; denote
Argument
1. From
as
c
sphericalpolygon
of its
Theorem
a
sides.
Given
sum
XXVI.
" 937.
all
A, draw
as
possible diagonals of the polygon,
2 sphericalA, I, II, etc.
forming n
Then, expressed in sphericaldegrees,
"
2.
AI
=
11=
A
A
.-.
.*.
area
in
"
(Z
AII
of
+
1578.
5 + Z
6) -180; etc.
r-(n2)180.
=
...
polygon
ABCD
" 54, 2.
" 309.
"""
180.
Prove
" 990.
Z3)-180;
+
Z
4 +
T-(n-2)
=
Ex.
I +
1 + Z2
(Z
2.
q.e.d.
Prop.
XXVI
by using
a
figure similar
to that used
216.
Ex.
1579.
92", 120",and
Ex.
to
Ex.
1580.
a
Find
the
140",in
Find
a
area
equivalentto six
Find
a
sphericalpolygon
sphere whose
the
angle of
sphericalpentagon
1581.
of
whose
radius
whose
angles are
80",
is 8 inches.
equilateralsphericaltriangleequivalent
angles are 90",100", 110",130",and 140".
an
angle of an equiangular spherical hexagon
sphericaltriangleseach with angles of 70".
equilateral
one
458
SOLID
GEOMETRY
Proposition
XXYII.
Theorem
If an isosceles triangle is revolved about a straight
line lying in its plane and
passing through its vertex but
not
intersecting its surface, the volume
of the solid generated
is equal to the product of the surface generated by
third
the base of the triangle and
one
of its altitude.
995.
Y
Fig.
isosceles A
Given
lying
in the
plane
the surface
by
volume
To
OE, and
passing through
of A
AOB,
of A
AOB-,
XF
as
a
str. line XY
and
O
let the volume
revolving about
AAOB
Fig. 3.
altitude
with
AOB
2.
secting
inter-
not
of the solid generated
axis be denoted
an
AOB.
volume
prove
I. If
is not
AB
AOB
=
and
IIXY
and
AC
2.
Prolong
3.
Then
4.
Volume
5.
.'.
to meet
BA
volume
AOB
FOB
volume
_L
BD
"
FOB
does
not meet
6.
But
7.
.-.
8.
But
9.
.-.
BD
volume
ttBD
volume
"
FO
FOB
"
BF
FOB
volume
FOB
+
FDB
=\tt ~BD2
as
=
7T
tt
area
area
volume
FOA.
BOB.
+
7T
"
=
twice
^
volume
"
\ BD2 DO
BDZ(FD+ DO) \ttBDFD
"
=
(Fig.1).
at F.
volume
=
XY
Only
XY
1
"
XY.
=
=
OE.
AB
Argument
1. Draw
1
area
area
BD
"
BF
of A
OE
FB.
FB
iOK
FOB
=
"
ttBD
BF
"
BF
BD
"
FO.
OE.
"
I OE.
by
BOOK
10.
Likewise
11.
.-.
volume
If
II.
The
See
Hint.
If
The
AOB
is not
AB
proof
III.
volume
The
996.
=
area
FB
area
Arg.
OE
AB
"
J
\
OE.
FA
area
"
FA)
i
"
Q#
Q.E.D.
lies in
XY
(Fig.2).
for the student.
" 995,
I.
(Fig.3).
is left
as
AOB
volume
=
student
exercise
an
may
:
for the
ACDB
student.
twice
"
volume
Outline
A.
of
of
a
a
sphere
sphere
Proof
of
(areaA'b'c' *-)$$
V=
=
=
A'E'
2
"
1
ttR
"
R
%7rR2"A'E'
("" 996, a
v
CO
.
(a) State and prove the corollaries on the volume
corresponding to "" 967 and 968.
the volume
on
(b) State and prove the theorem
corresponding to " 969.
1.
iOE
OE.
point A
10 of
"
area
"
exercise
an
or
i
"
=
and
FA
area
(areaFB
as
IIXY
AB
Volume
Hint.
9
FOA
=
IIXY
is left
Arg.
proof
=
459
IX
=
(areaAB
=
AE
=
f 7ra2XE
"
2
C
-n-a
"
and
967).
)\ a
a
-J"
"
"
and
("" 996, a
968).
F_|ir^""^J,_JZaA'E'
i?3
=
*f
~
TTcr
Proceed
4.
a
be
=
R
as
"
in
less
than
a3
XE
" 855, observing that
limit
(" 543, I), the
made
2
^1J"
any
since
the
limit
of
(" 593) ; i.e. R^^-a3 may
previously assigned value, however
of
a3=
R3
small.
(c) State,by
aid of
"
970, the
definition
of the
volume
of
a
460
SOLID
GEOMETRY
XXVIII.
Proposition
The
997.
of
the
surface
To
of its
sphere
by S, and
Given
of a sphere is equal
third
surface and one
volume
area
its radius
its volume
V=
prove
with
0
S
"
Theorem
"
by
denoted
to
the
its radius.
by R,
the
In
the
R.
semicircle
half of
a
number
the
by
about
AE
Then
3.
As
Denote
semiperimeter
as
an
axis
8'
of
polygon, of
by S',and
4.
5.
6.
7.
8.
9.
a
semi-
sides of the
is
ABODE
regular
half, is
limit.
S'
Also
limit.
approaches Sasa
And
a
approaches R as a limit.
.'. S'
a
approaches S R as a limit.
.'. "
\a approaches S i R as a limit.
But Vf is always equal to s' J a.
"
"
"
"
.
.'.
V=S-"R.
996, a.
the volume
repeatedlydoubled, v' approaches
as
"2. "
even
"i.
which
517, a.
it revolves
as
generated by
by V'.
ABODE
number
an
1. "
apothem
generated
solid
=
ABODE,
its
of the surface
area
r
the
inscribe
regular polygon with
of the
polygon
Reasons
ACE
of sides.
a, the
by
of its
area
V.
Argument
1.
product
Q.E.D.
V
3.
" 996,
c.
BOOK
998.
Cor.
the
D
denotes
If v
of
I.
diameter
other
Cor.
the cubes
as
diameters.
a
radii
spheres
two
and
to
are
tlie cubes
as
It is believed
Note.
middle
of the
that
fourth
of
the
each
of their
theorem
century
of in
already spoken
the volume
Find
of
" 972.)
mathematician
1588.
Ex.
the radius, and
R
I wDs.
=
volumes
the
early as
as
great Athenian
| Trie3
=
See
Historical
proved
the volume,
of their
(Hint.
1000.
was
The
n.
461
sphere,
a
V
999.
IX
sphere inscribed
a
809
in
999
"
by Eudoxus,
b.c.
""
of
and
a
896.
cube
whose
edge is 8 inches.
volume
The
1589.
Ex.
of
sphere is 1774"ir cubic centimeters.
a
Find
its surface.
Ex.
a
Ex.
the
and
Ex
those
metal
The
each
a
sphere equivalent to
of
a
sphere equivalent to
of the
water
theorem,
the
From
the
of
results
of
the
to
Ex.
a
1592
and
state, in
of
the
their
tude
alti-
sphere.
Place
empty
it into
fill the
Is the
with
cone
cylinder filled ?
the
form
of
a
the volume
of
a
corresponding
cone.
sphere : (a)
volume
the
Next
cylinder three times.
cylinder;(6)
cylinder with
to
these statements.
1001.
Historical
given
propertiesthat
a
cone
circumscribed
(Exs.
the
relation of the volume
circumscribed
Prove
it into
empty
1593.
equal
to 2
a
1590.
sphere in the cylinder,then fillthe cone with
cylinder. How
nearly is the cylinder filled ?
and
with altitude
cone
cylinder in the figure have
and
cone
in Ex.
cone
of
diameter
a
6.
the radius
as
of
B, the diameter
Ex.
to
of base
Find
1592.
water
to
the radius
dimensions
and
the
radius
1591.
same
the
Find
1590.
1592
thfeby
and
the
or
a
the
problem
given cylinder" (Exs.
volume
of
cylinder and
1593), are
author
The
Note.
sphere is
a
twice
due
himself
to
is
the
1590
two
volume
Archimedes.
spoken of
"
more
To
find
and
thirds
of the
The
a
sphere equivalent
1591), as well
of the volume
the
as
of the
corresponding
cone
importance attached
fullyin "" 542 and 974.
462
SOLID
Ex.
made
to hold
Ex.
the
A
1594.
|
Ex.
have
Ex.
Find
1597.
radius
of its inner
Ex.
1598.
A
is
volume
;
1002.
as
volume
of
weighs
93
16 inches
of the surface
area
the
volume
of the
and
the volume
sphere.
sphericalshell 5 inches
a
Find
pounds.
in diameter.
thick
if the
is 10 inches.
24 inches
sphere of
a
radius
of
half the
one
A
Defs.
by
circle
of
the
Find
of the howl.
the
volume
diameter
weighs 175 pounds.
weighing 50 pounds.
material
same
is B.
sphere
a
in
of the
Find
the
radius
of
sphere
a
given sphere ; twice
the
ume
vol-
the volume.
times
n
sphere
hemisphere is
exact
an
diameter
material
same
pine sphere
The
1599.
whose
the
the
is
in diameter
value.
surface
the diameter
Ex.
certain
a
numerical
surface
Find
12 inches
sphere
In
same
inner
sphere of the
a
1596.
the
Find
A
of
whose
gallon of water.
1595.
weight
bowl
CxEOMETRY
spherical
of
sector
a
a
is
sector
circle
a
solid closed
revolvingabout
a
figuregenerated
diameter
of the
axis.
an
C
Fig.
The
the
generated by
zone
of the
base
1003.
Def.
If
radius
one
sphericalsector is
sphericalsector is a
Thus
CD
as
base
called
(Fig.2).
the
by
zone
of
spherical
a
axis,arc
of the
of the circular sector
AB
AOB
will
sphericalsector
If circular
of the circular
sector
one
BC, will be
a
sector
generating
base, the sphericalsector
is
cone.
(Fig.1) revolves about
generate
a
which
zone
generated by circular
BOG
revolves
sphericalsector generated,whose
arc
is called
part of the axis, i.e. if the base of the
a
if circular sector
an
arc
3.
sphericalsector.
a
sometimes
the
Fig.
2
sphericalcone
about
base is the
(Fig.3).
diameter
will be the
sector
diameter
zone
AOB
CDy
generated
BOOK
1004.
Cor.
volume
product of
to the
equal
The
HI.
IX
of
base
Us
463
a
and
spherical
one
third
is
sector
the radius
of the sphere.
Outline
Proof
of
the volume
generated by polygon OA'b'c1,
v the
volume
generatedby polygon OABC, S the area of the surface
generatedby broken line A'b'c',s the
Let
denote
V
of
area
line ABC,
broken
generated by
surface
the
and
the
z
sphericalsector generatedby
the
sector
Then
V
S
=
"
i
R, and
ha
8
v
S
R
s
a
V
\a.
s
=
R2
S
But
cular
cir-
AOC.
S-"R
V
of
base
^?(Args.2-5, " 969).
=
-
a1
s
R2
V
R^R*
.
v
a
Then
2
a"
/-""
n
a
by steps similar
volume
of
the
" 996, b and
to
c, and
" 997, the
generated by circular
sphericalsector
Z"\R.
AOC=
1005.
Cor.
IV.
z
the
area
of
the
zone,
sector,
If v denotes the volume
of a spherical
tude
of the zone
forming its base, H the altiand
the
R
radius
of the sphere,
V=z-\R
=
Ex.
1600.
volume
of the
whose
Considering the earth
sphericalsector whose
altitude is
3
Compare
volume
from
your
1601.
of the
the
the
twijie
north
other
results with
those
Considering
as
a
base
sphere
is
a
radius
with
zone
B, find the
adjoining the
north
2 Tt
;
"
("979)
%vR'*H.
7?
pole and
("1004)
{H-2ttR)\R
=
Ex.
sector
the
"
"
Is the
one
volume
twice
the
other?
3
of Ex.
earth
1558.
as
a
sphere with
radius
R, find
the
base is a zone
spherical sector whose
extending : (a) 30"
volume
pole ; (ft)60" from the north pole. Is the one
?
Compare your results with those of Ex. 1559.
464
SOLID
1602.
Ex.
of
volume
the
results with
1603.
Ex.
of the
area
of the
What
included
between
This
Hint.
A
amount
of the outer
Ex.
16 inches.
A
Ex.
Find
1608.
volume
is 686
equal
".
to
Ex.
1610.
great circle have
of the
Ex.
from
edge
the
of small
of the
volume
bases
the
one
circles,
What
part
is that
tion
por-
equator.
of the
of the
pyramids
earth
radius
1603 ?
in Ex.
zone
and
sphericalsector.
a
in thickness
2 inches
3 inches
the
contains
is 6 inches.
Find
of
volume
thick
has
same
the radius
3, 4, and
are
of
"
sphere circumscribed
a
about
a
tangular
rec-
12.
sphere inscribed
a
of
diameter
outer
an
shell.
edges
a
the
of
in
sphere and the surface of
a
ratio of their volumes.
certain
a
whose
cube
same
numerical
and
value.
Find
a
cube
are
each
is the greater ?
Which
sphere the volume
How
bullets
many
sphere of lead 10 inches
is 10 inches
1006.
pole ; (") 30" and
equal ?
Compare your
of
the circumference
the surface and
a
the volume
sphere.
1611.
a
30" south
of the
surface
In
north
sphere with radius i?, find
a
of the
the volume
the
Find
the
the
centimeters.
The
1609.
Ex.
lying
i?, find
between
shell.
volume
Find
cubic
as
sphere whose
parallelopipedwhose
Ex.
volumes
of two
sphericalshell
the
the
radius
zone
?
planes
of the
Find
1607.
other
consists
a
a
the circumferences
zone
the
as
surface
1606.
earth
spherical shell
of material
from
part of the entire
volume
1605.
Ex.
two
are
the
is this
1604.
the
the
bases
equator,
entire surface
Ex.
"
whose
is
1560.
Considering
of the
30" north
of Ex.
zone
45"
with
sphere
a
base
and
Are
equator.
those
30"
(a)
:
as
whose
spherical sector
the
from
earth
Considering the
parallelsof latitude
45"
GEOMETRY
Defs.
\ of
in
inch in diameter
an
diameter
? from
a
cube
be
can
made
of lead whose
?
A
bounding surface
is
spherical wedge
consists
of
a
lune
solid closed
a
and
the
figurewhose
planesof
the sides
of the lune.
The
lune is called the
angle of
1007.
wedges
In
are
the lune
base
of
the
the angle of the
sphericalwedge,
and
the
spherical wedge.
the
Prove, by superposition,
following property
of
:
equal spheres,or
in
equal if their anglesare
the
same
equal.
sphere,two
sphericalwedges
466
SOLID
1010.
A
Defs.
GEOMETRY
is
spherical pyramid
whose
bounding surface consists of a
the planes of the sides of the spherical
polygon. The
spherical polygon is
the base, and the center
of the sphere
the vertex, of the sphericalpyramid.
solid
closed
figure
sphericalpolygon and
a
1011.
By comparison with " 957, b
" 958, prove the following property
of spherical
pyramids :
and
In
equal spheres,or
in the
sphere,
same
triangularsphericalpyramids
two
bases
symmetrical sphericaltrianglesare
are
1012.
Cor.
The
n.
volume
of
radius
Outline
of
equivalent.
spherical triangular
a
equal to the product of
of the sphere.
is
pyramid
the
whose
its base
and
third
one
Proof
1.
Pyramid O-AB'c' =c= pyramid
(y-A'BC (" 1011).
2. .-. pyramid O-ABC
+ pyramid
i R
2 A
O-AB'C1
=c=
wedge A
+ pyra("1009); pyramid O-ABC
mid
2 B
O-AB'C
wedge B
^ R;
+ pyramid O-ABC'
pyramid O-ABC
2 c
wedge C
\ R.
2 (A +
3. .-. twice pyramid O-ABC
+ hemispnere
i R
2 (A + B
4. .-. twice pyramid O-ABC
+ 360
5. .-. pyramid O-ABC
(A+ B + C"
180)J R
AABC
K
"\R
\R.
=
=
=
=
"
.
"
"
=
B
=
"
C)-|R.
c)i R.
-f
-f
=
=
=
1013.
is
in.
The
volume
of
1009
Ex.
1613.
and
Show
that the formula
of
spherical pyramid
any
equal to the product of its base
of the sphere. (Hint. Compare
Ex.
a
Cor.
and
with
" 997
q.e.d.
-
is a
one
third
the
dius
ra-
"805.)
specialcase
of
"" 1004,
1013.
1614.
In
a
sphericalpyramid
sphere whose
whose
base
inches,find the volume
trianglewith angles 70",80", and
radius
is
a
is 12
of
90".
BOOK
EXERCISES
MISCELLANEOUS
Ex.
is
equivalent to
of the
base
whose
wedge
a
volume
The
of
equiangular gon
pentaan
angle of the
an
is 30".
Find
sphericalpyramid whose
a
angles of 105" is 128
sphericaltrianglewith
the
is
base
angle
GEOMETRY
pyramid.
1616.
Ex.
SOLID
ON
sphericalpyramid whose
A
1615.
467
IX
base
cubic
ir
is
angular
equi-
an
Find
inches.
of the sphere.
radius
sphere whose radius is 10 inches, find the angle of a
base
has an
tude
altisphericalwedge equivalent to a sphericalsector whose
Ex.
In
1617.
a
of 12 inches.
Ex.
1618.
Find
the
depth of
The
altitude of
cubical tank
a
that will hold
100
gallons
of water.
Ex.
the
1619.
vertex
off is
must
half of the
one
weight of
as
mid-points
of the
What
A.
pyramid
Ex.
of the
Ex.
the
edges
the
ing
meet-
is the
ratio
answer
and
foot,find the
2 miles
long.
dozen,
or
oranges
averaging
?
the
result
base
and
D
as
1622
in calculus
proved
of
the
for
of tin
with
that
in order
that
a
cylindricaltin
est
given capacity may require the smallthe
the
base
its construction,
diameter
of
having
tin
of its base.
amount
consistent
of Ex.
rectangularparallelopiped? any parallelopiped?
a
capacity has its height equal
the
cubic
a
are
cube
a
can.
cylindricaltin
A
diameter
of
D
whole
equal the height of the
holding 1 quart ; 2 gallons.
to
nth ?
dozen.
cube
as
top and
possible amount
1625.
per
per
of the
must
Ex.
from
that the part cut
so
one
to
distance
what
pyramid.
It is
at
at 15 cents
30 cents
ABC
figureis
1624.
closed
can
of copper
figure,B, C, and
Is the
1623.
if the
same
base
inch in diameter
an
by plane BCD
Consider
vertex
the
third ?
one
pounds
part of the
off
cut
Hint.
the
550
\ of
at
1622.
at
?
pyramid
in diameter
In the
Ex.
the
inches
in diameter
inches
3^
At
Disregarding quality, and considering oranges as spheres,
is the
which
better
bargain, oranges
solids, determine
similar
averaging 2f
is H.
passed parallelto
whole
wire
copper
1621.
Ex.
i.e.
a
be
Allowing
1620.
Ex.
plane
a
pyramid
a
can
Another
Find
the
dimensions
of such
a
can
holding 2 gallonshas its height equal
cylindricaltin can with the same
to twice
required
the
for
the fact contained
diameter
making
in Ex.
of its base.
the
two
1624 ?
cans.
Find
the
Is your
468
SOLID
Ex.
1626.
A
is cast
in the
form
Ex.
1627.
form
of
of the
a
ball 12 inches
cannon
of
The
cube.
a
cube
Find
of Ex.
the diameter
cone,
GEOMETRY
in diameter
the
1626
of whose
of the
edge
is
is melted, and
melted, and
base
the lead
cube.
lead is cast in the
the
is 12 inches.
Find
the altitude
cone.
Ex.
Find
1628.
foot of iron
the cube
weight of the
ball in Ex.
cannon
if a cubic
1626
weighs 450 pounds.
The
1629.
Ex.
the
into
six
planes determined
by the diagonals of
cube
a
divide
equal pyramids.
Ex.
1630.
Let
Ex.
1631.
In
and
mid-points of VA, AB, BC,
and
Prove
DEFG
a
CV, respectively,of triangular pyramid V-ABC.
parallelogram.
allel to
of
a
parallelto
points of
the
other
Hint.
Ex.
oppositeedges of
and
hence
Draw
DF
is
4f
inches
at
cubic
Ex.
the
inches
Ex.
a
of ice
Find
total
^^^Ijr*"*
,"7\tr
F^****JrE
bisect
"
these
two
6| inches high ;
top, and
the
at
actually
packed
about
of
area
required lines
of the
two-quart
can
be
ice
cream
the
tub
9f
hold
the
2
to
lateral faces
The
buildingon
was
of
moves
foundation
the
Broadway,
the sand
filled with
of concrete
and
a
triangular pyramid
work
York
New
quicksand
1637.
1638.
concrete
How
(6)
tude
alti-
whose
to
of the
the total
Woolworth
City, it was
bed rock, to
equilateral
are
Find
area.
55-
Building, a
necessary,
sink the
A
From
Find
In Ex.
to
within
feet of the
30
in order
that
caissons
surface, how
required,considering 1 cubic yard
were
mid-point of AB.
Ex.
quarts ?
the
feet.
Ex.
inches
instances,of 131
huge shafts of concrete to a depth, in some
If the largestcircular caisson, 19 feet in diameter, is 130 feet deep
contain
loads
In
1636.
penetrate
and
6| inches in
high, inside
is
?
can
and the altitude of the pyramid is 6 inches.
triangles,
story
freezer, the
regular tetrahedron
a
?
centimeters.
1635.
Ex.
I \
point.
inches
can
the
tetrahedron
\
yS
mid-
joining the
Are
the
Si~~"\A
by a plane
parallelogram.
a
l\
yr
r
made
and
Does
(a)
a
yt
par-
that any
Mountain
8
bottom,
1634.
is
White
a
a
EG.
in diameter
measurements,
many
in
meet
and
In
the
Prove
lines
three
The
1633.
diameter
edge VB?
to
opposite edges is
two
G be
figure,is plane DEFQ
triangularpyramid
1632.
Ex.
can
the
edge AC?
section
each
D, E, F,
draw
the
a
line
meeting line XFin
locus of C
1637, let XY
arbitrarilyin plane XY.
be
as
a
B
moves
to
B
a
many
load ?
the
; let G be
in line XY.
plane. Find
the locus of C
as
B
is twice
its altitude
the
pyramid
the
base
in
railway
a
being
is it not
Why
the
section
a
to
building.
The
at
the
allowed
to
feet to
cubic
top, and
a
sphere
cube
;
Ex.
Ex.
(c) the
a
The
of the
shown
of 10
inches.
It
semicircle
Ex.
top, and
at
divides
Ex.
section
of
each
A
5^
inches
of
dimensions
a
"7"
12
16
deep, how
of
room
the
at
If the
of
in the form
tower
frustum
a
certain
a
public
ameter
feet in di-
bottom,
12
in the
tank
water
is
cubic feet of water
many
? how
never
would
barrels, counting i\
many
1647.
Find
the
(a)
:
the
of
to
the
same
the
and
:
(a)
cone
house
3 feet
each
of
the
the^plane. Find
altitude
the
to
volume
of
the
the
as
sphere
;
cylinder in
(")
the
to
a
s\
denoting
rectangle with a
yj
Find
the
total
cups
to
at the
a
two
TTTN
"L/
"*
in
radiating surface
(Answer
B,
inches
How
bottom.
and
supplied
was
room
on
in diameter
many
solids
the
same
cups
whole
to nearest
center
the
of
a
room.
at
the
of coffee
number.)
parallelopiped
equal ?
side of
plane AB,
plane AB, making equal angles with
the locus of point O.
(Hint. See Ex. 1237.)
to
point
0
a
2
deep, 4J
quart ?
points, P
for
gl
is
plane passing through the
equivalent solids. Are these
drawn
the
high.
radiator
a
in diameter
From
cylinder,and
of the
ratio
heat
Any
1648.
a
cube.
base
coffee pot is 5 inches
it into two
in
cylinder; (b) the ratio of the cylinder to
numbers
end.
allowing 6
are
of loads
number
?
deep.
steam-heated
figure,the
1646.
lines
the
cube.
radiators
consists
will it hold,
Ex.
a
cross
in the
ratio
the
In
average
feet
sphere
has
cone
Find
series
;
of
plane parallelto
a
cut.
diameter
cube.
a
ratio
A
1645.
in
of the
cylinder; (c) to the
Ex.
the
emergency
that
1644.
1643.
4
total surface
of
name
sphere with radius B is inscribed
A
to
an
are
barrel ?
cylinder is inscribed
the
in
14 feet
of
case
1643.
Ex.
16
less than
in
load, find
a
in the
placed
was
is 16 feet
tank
get
available
cone
bases
figure,
the
pyramid
a
by
protectionagainst fire,a tank
For
right circular
a
of
made
the
length, the average
in
represented in
as
Find
(a)
square
42" cubic feet
is
pyramid
a
base.
| mile
cut
frustum
a
1642.
Ex.
two
of
area
numbers
the
by
base,
denoting feet. Give the
geometrical solid represented by
the
be
the
Allowing'l cubic yard
section
cross
the
from
foot
one
1641.
earth
of
of
and
find
regular square
a
of
side
one
(6)
;
of
volume
The
1640.
Ex.
of
frustum
a
If cubic feet of granite; the edges of the
\\ feet, respectively. Find the height of the shaft.
and
Ex.
of
form
the
16
pyramid contains
feet
in
granite shaft
A
1639.
Ex.
469
IX
BOOK
in
470
GEOMETRY
SOLID
A
1649.
Ex.
pyramid.
square
bases
feet
12
are
is 8
inches, if
1651.
Ex.
the
inches.
height of
the
cubic
feet of material
edge
of the
1
The
The
main
5 feet
does
the
that
lies in the
of
the
6 inches.
part of the
view
a
of the top of
the volume
Hint.
a
the
stone
Ex.
righttriangularprism
1652.
the form
of
The
a
wasted
were
Ex.
main
part
Ex.
Ex.
tank
of
view
a
3.
3 represents
Fig.
late
Calcu-
above.
and
In
part of the
stone
sists
con-
two
righttriangularprisms B, and
sists
Fig. 3 shows that the top conrectangularpyramids.
1651
from
cut
was
cubic
many
a
solid rock
feet of
in
granite
cutting?
the
of Ex.
monument
rock
a
being polished.
polished surface.
1651
finish,the
Find
the
the
front
of the
ends
two
and
rear
surfaces
of
square
feet
number
main
of the
of
rock
of
1654.
circle whose
lateral
total
four
in Ex.
monument
the top, have
finish and
the
8
above.
main
the
rectangularparallelopiped. How
in the
1653.
part, and
whose
of the stone.
From
a
a
numbers
high,
looking directlyfrom
that
Fig. 2 it is seen
of a rectangularparallelopipedA,
rectangular pyramid at each corner.
of
from
pyramid.
the
20
looking directlyfrom
stone
of the
of its
square
side of
one
lar
regu-
contain?
cut
Fig. 2 represents
Fig.
main
be
is 5 feet
stone
feet
chimney
base
a
edges
6
can
and
of
the
flue is
,8
the
and
granite monument,
a
part
frustum
a
high,
largestcube
cube
represents
being
stone
of
altitude is 10 inches
whose
face of the
one
Fig.
denoting
feet,respectively.
many
pyramid
feet
is 120
chimney
8
Find
1650.
regular square
base
and
How
throughout.
Ex.
The
form
is in the
factory chimney
area
1655.
in Ex.
The
radius
of the
base
of
is
and
B,
a
regular pyramid
is
the altitude of the
a
triangleinscribed
pyramid
is 2B.
Find
in
a
the
pyramid.
Find
1323, if
the
a
weight
cubic
in
pounds
foot of water
of the
water
weighs
1000
required to
ounces.
fillthe
BOOK
of the
and
BOC
the
about
on
solid
a
in diameter
cut
axis.
Check
your
of the
of the
the
Ex.
1661.
Four
spheres,each
the
others.
Ex.
with
Find
radius
of 6
area
total
the
total
the
inches, if
=
"""
the
height of
there
=
"""
B
=
=
in
of
angles
=
E
=
inches,are
one
placed
to each
of
triangularpileof spheres, each
layers; four layers; n layers.
GEOMETRY
given
in "
altitude
of
761, the
ing
follow-
circumference
=
=
area
=
area
=
number
of
and
or
of
of frustum
of
a
sphere.
of a
spherical excess
sphericaltriangle.
degrees in
lune
a
or
of base
base
area
of surface
T
sum
of the
Z
area
wedge,
in general, of
of
frustum
of
base of frustum
of
sphere,
angles of a spherical
polygon,
volume
:
angle
of.sphere,
or
of upper
S:
W:
the
cone,
cone.
diameter
spherical
sphericaltriangle or
of
cone,
of
of upper
or
of lune.
radius
cone,
circumference
a
lower
of frustum
base
of
radius
of base
general
zone
sphericalpolygon,
N
tor,
sphericalsec-
base
D
K
of
of
a
1658.
sector,
L
of
=
H
polygon.
lower
c
left in Ex.
of 6
radius
a
sphericalpolygon
sides of a spherical
in
by
being tangent
triangularpile.
notation
degrees
wedge,
pyramid.
C
1658
SOLID
to the
of
number
base
a
three
are
.
b, c,
ume
vol-
:
the
a,
is 12.
spherical
Ex.
with
OF
In addition
will be used
A, B, C,
the
sphericalsurface
height of
FORMULAS
1014.
edge
bored
was
triangularpile,each
a
Find
1662.
the volume
whose
part left.
Find
in
for
result
1660.
surface
edge
1544,find
the volume
is 12.
revolving isosceles A
generated by
volume
plane
in Ex.
Find
consists of two
out
Ex.
a
whose
regular tetrahedron
in diameter.
an
X7as
findingthe
1543, find
out.
part
1659.
Ex.
in Ex.
obtained
formula
about
10 inches
cut
part
The
Hint.
obtained
regular tetrahedron
a
6 inches
hole
A
sphere
a
of the
cones
circumscribed
1658.
through
in
using the
By
sphere
Ex.
inscribed
sphere
1657.
Ex.
of the
By using the formula
1656.
Ex.
471
IX
of
of
a
a
a
wedge,
zone.
472
SOLID
GEOMETRY
FIGURE
FORMULA
REFERENCE
I
Prism.
8
Right prism.
8=P-
Regular pyramid.
S=%P.L.
of
Frustum
P
=
E.
"762.
'
H.
" 763.
"766.
regular pyramid.
"767.
Rectangular parallelopiped.
V=a-b-c.
"778.
Cube.
V=E\
"779.
Rectangular parallelopipeds.
V
a-b-c
"780.
Rectangular parallelopiped.
"782.
Rectangular parallelopipeds.
"783.
Any parallelopiped.
"790.
Parallelopipeds.
"792.
Triangular prism.
"797.
Any
prism.
"799.
Prisms.
"801.
Triangular pyramid.
"804.
Any pyramid.
"805.
Pyramids.
V
Similar
tetrahedrons.
B'
of any
Truncated
pyramid.
cylindersof revolution.
V=
\B(E
cones
of revolution.
"815.
+
"817.
+
E").
C-H.
"858.
8
=
2wBH.
" 859.
T=2ttB(H+B).
"859.
H*
S_=
_B*
B'2'
H'2
8
=
\C-L.
8
=
w
"
_
H'2~
"864.
"873.
BL.
H*
__
8'
"864.
=BT* _B*
T=ttB(L
Similar
b+y/Tn").
E'
=
T'
cone.
+
8
T
Right circular
"812.
E's'
V=}H(B
righttriangularprism.
Right circular cylinder.
Similar
"807.
i
V_=E^
V
Frustum
H
.
"875.
+
B).
Z2
._
B*_
L'2~ B'2'
"875.
"878.
474
SOLID
APPENDIX
GEOMETRY
TO
SPHERICAL
1015.
A
Defs.
whose
bounding
GEOMETRY
SOLID
SEGMENTS
is
spherical segment
surface
consists
of
a
closed
solid
a
zone
and
two
figure
parallel
planes.
of Two
Spherical Segment
called the
1016.
planes
sphere formed by the two parallel
of the spherical
segment.
bases
I.
Proposition
1017.
To
derive
segment
in
a
terms
of
Given
and
To
an
the
Problem
for
formula
its altitude.
as
Base
State,by aid of "" 976 and 977, definitions of:
of a sphericalsegment.
(6) Segment of one base.
Defs.
(a) Altitude
EF
of One
sections of the
The
are
SphericalSegment
Bases
the
the
radii
volume
of
of
its
a
bases
ical
spherand
E
sphericalsegment generated by ABCD
revolvingabout
denoted
axis,with its volume
by r, its altitude by h,
radii of its bases by rx and r2, respectively.
derive
a
formula
for
V
in terms
of i\, r2, and
h.
477
APPENDIX
.
pyramid
\
the
.'.
V=\H"B
substitutingin
By
(a) the volume of
pyramid (" 805) ; (c) the volume
for
formula
:
Solve
1666.
Ex.
1651
Ex.
The
1022.
(a) Any
a
ratio
same
+
\H
=
4 j%.
"
4:M).
Q.E.F.
prismatoid formula, derive
prism (" 799) ; (6) the volume
a
frustum
of
the
by applying
1023.
other
any
the
of
a
pyramid (" 815).
a
prismatoid
formula
to
the squares
as
the squares
as
The
Def.
of any
surfaces of
is the ratio of any
If
two
of
of
ratio
two
two
following:
similar
two
similar
two
homologous edges.
similar
any
two
similitude
are
are
Def.
homologous edges.
of
^"^^^/\
q"
_
_
_
~
~
~
~
ob
oc
1667.
Ex.
point 0
*
Construct
lies between
See
" 811.
the
CD
0
and
"""
a'b'c'd'
...
A', b',C',d'}etc.,
to
C'
!
\
E'
od
polyhedrons are
two
drons
polyhe-
^
qif
_
oa
similar
two
^^^7\--^/\
that
q"
oa[
the
homologous edges.
two
A, B, C, D, etc.,in such
manner
polyhedrons have
polyhedrons have
polyhedrons AB
situated that lines from a point
so
divided
by points
g'
1024.
polyhedrons have
homologous edges.
two
homologous faces of
total
ratio
same
as
ratio
same
the
prove
homologous edges of
two
(c) The
the
"
the
of
should
student
two
(b) Any
a
=
\H mj
\ H"4tM.
POLYHEDRONS*
SIMILAR
the
+
"
part of the monument.
each
the
} II mx
pyramids
mx+
"
\H-b+\H."M="rH(B+b
+
1665.
JH
=
of all lateral
volume
.*.
Rs.
P-ADC
two
two
said to be radially placed.
polyhedrons radially placed and so that
polyhedrons ; within the two polyhedrons.
In this discussion
only
convex
polyhedrons
will be considered.
478
GEOMETRY
SOLID
III.
Proposition
1025.
similar.
Given
to
point
To
two
radially placed polyhedrons are
Any
(See Fig. 2 below.)
polyhedronsEC and 'E'c' radiallyplaced with respect
0.
polyhedron EC
prove
AB, BC, CD, and
DA
are
~
polyhedron
E'c'.
a'b',B'c',c'd\ and
IIrespectively
to
" 415.
D'A'.
.-.
Theorem
ABCD
and
IIA'B'C'D',
is similar
" 756, II.
to it.
to the corresponding
polyhedron EC is
face of polyhedron E'c',and the faces are similarly
placed.
Again, face All IIface a'h',and face AF IIface A'F'.
each
Likewise
.-.
dihedral
Z
face of
AE
each
Likewise
dihedral
=
dihedral
corresponding dihedral
.-.
.-.
~
Z
each
Z
polyhedron EC is equal to its
of polyhedron E'c'.
sponding
polyhedron EC is equal to its correZ
of
polyhedralZ of
polyhedralZ of polyhedron E'c'.
polyhedron E'c'. "811.
polyhedron EC
" 18.
q.e.d.
~
Proposition
1026.
Any
two
similar
IV.
prove
may
be
radially
q'
Fig.
Given
Theorem
polyhedrons
placed.
To
A'e'.
two
similar
that
Fig.
i.
XM
polyhedronsXM
and
E'c' may
be
and
2.
E'".
radiallyplaced.
479
APPENDIX
Outline
Take
1.
polyhedron
and
EC
that
so
point
any
2.
Then
3.
Prove
OA'
:
polyhedron E'C? and construct
is radially
placed with respect to E'C1
0A
OB'
=
:
polyhedronEC
that
to
respectively,
the
that
Prove
OB
A'B'
=
""
:
KL.
polyhedron E'c'.
~
A
A
dihedral
of
polyhedron EC
polyhedron XM,
of
A
polyhedron
EC
of
6.
.-.
7.
But
8.
.-.
and
XM
E'c'
placed in the positionof
radiallyplaced.
be radiallyplaced.
are
V.
Proposition
1027.
base
is cut
pyramid
a
EC.
q.e.d.
Theorem
by
plane parallel
a
to
its
:
The
I.
II.
of
If
XM.
=
be
may
E'c' may
and
XM
each
equal,respectively,
are
5.
EC
equal,
being
are
polyhedron E'c'.
polyhedron XM.
that polyhedron EC
Prove, by superposition,
polyhedron
" 1025.
of
the dihedral
the faces of
t6 the faces
=
dihedral
the
to
equal,respectively,
4.
within
0
that it
so
Proof
of
any
pyramid
The
two
similar
off is
cut
to the
to each
are
pyramids
homologous edges.
two
given
other
as
mid.
pyra-
the
cubes
O
The
Hint.
proofs are
For
the
left
as
exercises
proof of II, pass
B'D', B'E', etc., dividingeach
Then
of
pyramid
pyramid O-BCD
O-E'B'D', etc. Use " 812 and
~
for the student.
planes through OB'
the
pyramids
O-B'C'D'
a
method
;
into
and
diagonals
triangularpyramids.
pyramid
O-EBD
similar to that used
~
in
mid
pyra-
" 505.
480
SOLID
GEOMETRY
Proposition
1028.
cubes
the
similar
Two
of
any
VI.
Theorem.
each
to
polyhedrons are
homologous edges.
two
other
as
C
similar
polyhedrons
and
XM
E'c', with
their volumes
denoted
by V
and with KL
and V',respectively,
two
Given
A'b' two
and
homol.
edges.
:3
KL
To
prove
A'B1
and E'C} are radially
positionEC, so that XM
placed with respect to point 0 within both polyhedrons. " 1026.
of pyramids O-ABCD,
Denote
the volumes
etc.,by
O-AEFB,
of pyramids O-A'b'c'd',
O-A'e'f'b',
vx, v2, etc.,and the volumes
etc.
etc.,by v-iy v2',
Place
XM
in
-^
^
;
"1027,11.
etc-
=3;
=
AB
j%,=
("1022, a);
-
A^B'
AB
AB
;i
A'E'3
"401.
_
"k'
polyhedron EC
polyhedron E'C'
1029.
the teacher
IX
Note.
will find
Vi + V
IB1
Since
AB3
KL3
V
.
ArB'
4-
Q.E.D.
__
__
^B13
" 1028
plenty of
'
was
V'
assumed
exercises
this principle.
illustrating
J^3
early in
throughout
(see "814),
VII, VIII, and
the text
Books
INDEX
(The
numbers
refer to
articles.)
Art.
Acute
Adjacent
Adjacent
42
dihedral
angles
.
.
Alternation
of
of
of
of
of
of
of
of
of
cone
cylinder
parallelogram
prism
prismatoid
pyramid
spherical segment
trapezoid
triangle
....
....
.
.
zone
Analysis, of
a
68
Angles, complementary
182
corresponding
designation of
39, 50, 667, 696
...
angles
Altitude, of
Art.
98
triangle
proportion
.
.
.671
396
842
825
228
733
1020
751
1016
229
100
976
397
.
difference
exterior
homologous
.
.
vertical
Antecedents
Apothem
Application
Arc
degree of
length of
major
minor
Arcs, similar
Area, of a figure
of circle
of rectangle
of surface
.
.
.
.
.
adjacent
exterior
alternate interior
alternate
'
42
....
.
.
.
182
.182
44
......
supplementary-adjacent
...
.
51
86
supplementary
....
.
417, 424
86, 182
.
of
sum
.
.
.
oblique
of a polygon
subtracted
....
Argument
only
Arms, of isosceles triangle
of right triangle
Assumptions
Axiom
Axioms
note
on
Axis, of circle of sphere
of right circular cone
Base, of
of
of
of
of
of
.
.
43
69
72
70
384
533
79
121, 280
296
551
291
291
565
471
558
468-471
467
171
94
96
12, 54
55
54
600
906
844
"
"
.
.
.
.
.
.
cone
isosceles triangle
polygon
pyramid
spherical pyramid
spherical sector
Bases, of cylinder
of prism
of spherical segment
of
....
.
.
.
.
.
.
.
.
zone
Birectangular sphericaltriangle
.
481
182
182
.
interior
....
.
87,
exterior interior
.
,
44
44
equal
of constructions
151, 152, 176 (Ex. 187) 274, 499
274
of exercises
38
Angle
47
acute
of circle
292
at center
of regular polygon
534
at center
bisector of
53, 127, 599
central
292
71
degree of
dihedral
666
inscribed in circle
363
inscribed in segment
364
magnitude of
41, 669
obtuse
48
of lune
983
of one
71
degree
of spherical wedge
1006
of two intersecting arcs
916
.692
polyhedral
reflex
49
right
45, 71
.199
right and left side of
sides of
38
solid
692
917
spherical
69
straight
tetrahedral
698
trihedral
698
of
38
vertex
38
Angles
added
43
....
of
840
94
99
748
1010
1002
822
727
1015
975
952
482
INDEX
Art.
.53, 127, 599
of line
52, 145
Bisectors of angles of triangle
126
Bisector, of angle
.
.
.
.
.
.
,
.
.
Center, of circle
of regular polygon
of sphere
119, 278
....
.
Center-line
Central
angle
.
.
...
Chord
Circle
arc
of
of
of
chord
of
circumference
center
of
.
circumscribed
determined
diameter
escribed
.
.
Centroid
area
.531
901
325
292
265
281
119, 276
121, 280
558
119, 278
281
119, 277, 550
300
324
.
.
of
282
322
inscribed
317
radius of
119, 278
of
secant
285
of
sector
287, 564
of
288
segment
286
to
tangent
326
Circles,concentric
330
tangent, externally
330
tangent, internally
841
Circular cone
Circular
831
cylinder
Circumference
119, 277, 550
550, 552
length of
of
540
measurement
circle
300
Circumscribed
317
Circumscribed
polygon
926
Circumscribed
polyhedron
852
Circumscribed
prism
868
Circumscribed
pyramid
Circumscribed
929
sphere
motion
86
Clockwise
Closed figure
83, 714, 715, 934, 935
82
Closed line
17
Coincidence
Collinear
28
segments
Commensurable
quantities
337, 342, 469
Common
337
measure
31
Compasses, use of
68
Complementary*
angles
.171
Complete demonstration
398
Composition
400
and division
Composition
Concentric
circles
326
57
Conclusion
lines
196
Concurrent
Cone
839
altitude of
842
.....
.
.
.
.
.
.
.
.
....
....
.
.
.
.
.
.
.
....
.
....
.....
.
....
...
.
.
.
.
Art.
840
841
840
879
840
843
876
866
843
Cone, base of
circular
element
of
frustum
of
lateral surface
of
....
oblique
of revolution
plane tangent
right circular
spherical
to
1003
of
volume
of
Cones, similar
Conical
surface
directrix of
element
of
840
vertex
892,6
generatrix of
lower
upper
nappe
nappe
of
of
.
.
.
of
vertex
Consequents
Constant
Construction
of triangles
Continued
Converse
proportion
....
theorem
Convex
877
837
838
838
838
838
838
838
384
346, 347
123
269
405
135
697
.
polyhedral angle
Coplanar points, lines,planes
Corollary
.
Counter-clockwise
Cube
.
.
.
.
motion
.
Cylinder
altitude of
bases of
circular
element
of
lateral surface
of
....
oblique
of revolution
plane tangent to
right
right circular
right section of
-
.
.
668
58
50
742
821
825
822
831
822
822
824
861
850
823
_
832
830
888
863
819
.820
820
820
volume
of
Cylinders, similar
Cylindrical surface
directrix of
element
of
generatrix of
Definition, test of
Degree, of angle
of
arc
spherical
Demonstration
complete
Determined
Determined
circle
line and
Determined
plane
point
.
138
71
296
986
56, 79
171
324
25, 26
608
484
INDEX
Art.
Historical
notes
Lambert
Lindemann
of Alexandria
Menelaus
Metius
of Holland
Morse, S. F. B
Plato
.
569
569
992
569
520
542
474
268, 542, 787
371
Newton
Origin of geometry
.
..
.
.
.
.
.
....
....
Plutarch
114
asinorum
Pons
Ptolemy
569
.
Richter
Seven
Wise Men
Shanks
Snell
Socrates
Squaring the circle
Stobseus
Thales
Zeno
parts
.
.
.
569
.371
569
946
.
787
569
114
371, 510
354
.
.
....
109, 110, 417,418,424
Hypotenuse
Hypothesis
96
57
..
.
Icosahedron
Inclination
of line to plane
Incommensurable
quantities
.
719
665
.
.
Inscribed
Inscribed
polygon
polyhedron
Inscribed
prism
Inscribed
pyramid
Inscribed
sphere
Instruments, use of
of two
.
.
363
317
299
.928
851
867
927
31
surfaces
Isoperimetric figures
trapezoid
triangle
.
.
614
395
572
227
94
94
94
94
94
.
....
Isosceles
Isosceles
of
base of
sides of
arms
vertex
.
angle of
.
.
.
.
.
closed
82
curved
29
25
406
434
determined
divided
divided
divided
externally
harmonically
....
.
.
....
.
.
area,
of frustum
.
of
of
of
of
prism
pyramid
right circular cone
right circular cylinder
Lateral edges, of prism
of pyramid
Lateral faces, of prism
....
pyramid
..,,,,
.
.
.
.
of
.
.
.
.
.
760
760
760
872
857
727
748
727
748
.
.
.
.
.
.
.
.
....
.
.
.
144
finding of a
of a point
of all points
problem, solution
130
130
of
.
143, 60,1
982
983
983
983
Lune
Lateral
.
and
in
extreme
ratio
mean
464, 465 (Ex. 763), 526
406
divided internally
inclination
of
665
630
oblique to plane
of centers
325
629
parallel to plane
619
perpendicular to plane
656
projection of
23
right
of
27, 406
segments
straight
21, 23
.921
tangent to sphere
27
Line segment
Lines
20
196
concurrent
31
difference of
407
divided proportionally
homologous
109, 110, 418, 424
46
oblique
177
parallel
45
perpendicular
product of
425, 475, 476, 511
rectangle of 425, 475, 476, 511
Locus
129, 130, 131, 601
.130
an
assemblage
as
130
a
path
as
...
.
339, 343, 344, 470
159, 161
Indirect proof
Inscribed
angle
Inscribed
circle
Intersection
Inversion
.
344,510,514,723,787
Pythagoras
Homologous
Art.
Lateral surface, of cone
840
of cylinder
822
of frustum
of pyramid
.760
,m
551
Length, of arc
of circumference
550, 552
of perpendicular
166
of secant
426
of tangent
319
Limits
346, 349-351, 590-594
Line
4, 7, 11, 27
bisected
52, 145
broken
30
angle of
sides of
vertices
Magnitude
Major arc
of
of
an
angle
Maximum
Mean
proportional
Means
.
.
.
41, 669
.
.
.
291
571
.386
-385
485
INDEX
Art.
Measure,
common
numerical
unit of
Measurement
of angle
of arc
of circle
of circumference
of distances
by
.
.
.
.
335, 467,
335,
.
.
.
.
.
.
540,
540,
.
.
triangles
of line
of rectangle
of surface
Plane, tangent
115
Point
determined
of tangency
of
means
Art.
337
770
466
335
358
358
558
552
335, 595
tangent
tangent
Plane
Plane
Plane
to
to
to
cone
866
850
....
cylinder
sphere
angle of dihedral
figure
....
.
.
.921
.
angle
670
36
36
631
672
615
.
geometry
Planes, parallel
perpendicular
postulate of
5, 6,
....
.
.
.
.
.
.
.
.
-
.
....
.
.
"
"
.
.
.
.
""
.
.
.
.
.
.
.
...
.....
.......
.
.
.
.
.
....
.
....
'
11
26
468-471
286
466
Polar distance
of circle
.911
Polar triangle
943
Measure-number
Poles of circle
907
335, 467, 471, 595, 770
84
.251
Polygon
Median, of trapeze id
86
102
of triangle
angles of
base of
265
99
Median
of triangle
center
317
circumscribed
Methods
of attack
88
diagonal of
103, 104, 106, 110 (Ex. 62),
90
equiangular
115, 151, 152, 158, 159, 176
89
equilateral
(Ex. 187), 269-274, 275, 397,
87
exterior angles of
398, 425, 499, 502, 513
inscribed
299
571
Minimum
85
291
perimeter of
Minor
arc
91, 515
regular
84
sides of
and lower
746, 838
Nappes, upper
Numerical
335, 467, 770
spherical_-^_-.v__!__i_, .936
measure
84
^.-verticesof
51 "Polygons, mutually
equiangular 417
Oblique angles
.418
sides proportional
843
Oblique cone
419
similar
824
Oblique cylinder
692
46
Polyhedral angle
Oblique lines
697
97
Obtuse
convex
triangle
of
694
719
dihedral
angles
Octahedron
694
One
edges of
to one
correspondence
595, 596
694
of
element
136
Opposite theorem
694
face angles of
81
Optical illusions
694
faces of
171
Outline of proof
695
:
parts of
693
of
177
vertex
Parallel lines
707
631
Parallel planes
Polyhedral angles, symmetrical
708
vertical
220
Parallelogram
716
228
Polyhedron
altitude of
about
926
circumscribed
sphere
228
base of
718
diagonal of
223, 243
Parallelograms classified
717
739
edges of
Parallelopiped
717
of
741
faces
rectangular
928
inscribed
in sphere
740
right
720
92
regular
Pentagon
717
vertices of
85
Perimeter
of polygon
1024
672
radially
placed
Polyhedrons,
Perpendicular
45, 166, 619, 620,
similar
621
811, 1022
foot of
114
asinorum
Pons
672
Perpendicular planes
Portraits
554
Pi
,
542
Archimedes
568
evaluation
of
237
Descartes
Plane
32, 33, 34
114
Euclid
determined
608
520
Gauss
line
620
to
perpendicular
'*
.
.
"
.
486
INDEX
Art.
Portraits
Plato
787
510
371
15
122
615
178, 179
40, 54, 606
24, 54
16, 54
726
733
727
Pythagoras
Thales
Postulate
circle
.
of planes
parallel line
....
revolution
...
straightline
transference
Prism
altitude of
bases
Art.
Proportion, extremes
of
about
circumscribed
cylinder 852
.851
inscribed in cylinder
.
.
of
lateral area
lateral edges of
lateral faces of
oblique
quadrangular
regular
right
right section of
triangular
truncated
Prismatic
surface
.
Prismatoid
altitude of
Problem
defined
of computation
Problems
Problems
of construction
analysis of
151, 152, 176,
...
.
.
.
.
760
727
727
731
732
730
729
728
732
736
724
1019
1020
59
59
59
.
of
.
of
of
of writing
means
terms
.......
ways
....
mean
third
simplified
.
.
altitude of
base
of
circumscribed
about
frustum
of
inscribed in cone
lateral area
of
lateral edges of
lateral faces of
cone
....
quadrangular
regular
.
slant
height
.
*
....
Proportional, fourth
Proportions
Proposition
Pyramid
385
385
383
382
387
386
386
389, 397
56
747
751
748
868
754
867
760
748
748
749
752
764
1010
749, 750
753
748
744
.
of
spherical
triangular
truncated
of
vertex
Pyramidal surface
.
.
Quadrangular prism
....
Quadrangular
pyramid
Quadrant
Quadrilateral
Quadrilaterals classified
.
.
.
732
749
295
92
220, 243
.
.
1024
Radially placed polyhedrons
Radius, of circle
119, 278
532
of regular polygon
of sphere
901
Ratio
340-343, 382, 384
.
(Ex. 187), 274, 499
discussion
of
123
.
.
.
....
proof of
123
123
476, 511
451, 656
450, 655
79, 123
275
solution of
Product
of lines 425, 475,
Projection, of line
of point
Proof
of
analytic method
by exclusion
159, 161
275
by successive substitutions
.
.
.
...
.
....
indirect
.
.
.
.
.
159, 161
necessity for
81
171
outline
of
reductio ad
absurdum
of
synthetic method
Proportion
analysis of
.
159, 161
.
.
antecedents
of
by alternation
by
by
by
by
composition
composition
division
inversion
consequents
continued
of
.,.,,,
and
division
.
275
382
397
384
396
398
400
399
395
384
405
....
antecedent
consequent
of
of
and
extreme
384
384
mean
....
(Ex. 763), 526
464, 465
of
of
of
of
of
of
surfaces
two
any
similitude
commensurables
incommensurables
two
magnitudes
two
solids
.
.
of
lines
of two
Reductio
425, 475,
parallelopiped
figure
ad
absurdum
Regular polygon
angle at center
of
apothem
center
radius
Regular
.
.
.
of
.
.
,
.
.
.
...
of
of
polyhedron
.
.
area
Rectilinear
342
343
.341
775
223
468-471
476, 511
-741
37
159, 161
91, 515
534
533
531
532
720
,
,
.
Rectangle
Rectangular
.472
418, 1023
two
two
.
.
....
.
INDEX
487
Art.
730
752
Regular prism
Regular pyramid
variables
Related
Rhomboid
.
352, 594
.
.
224
226
Rhombus
Right circular
cone
axis of
of
lateral area
slant height of
Right circular cylinder
of
lateral area
.
Right
Right
Right
Right
.
.
cylinder
parallelopiped
prism
section, of cylinder
of prism
Right triangle
....
.
.
.
843
844
872
865
832
857
823
740
729
830
728
96
96
96
96
of
arms
of
hypotenuse
sides of
Ruler,
of
use
31
......
Scale, drawing to
Scalene
triangle
Secant
length
of
Sector
spherical
of circle
of line
of sphere
Segments, added
collinear
difference of
of a line
similar
Segment,
subtracted
sum
of
Semicircle
Semicircumference
Series of equal ratios
....
.
.
Sides,of angle
of polygon
Similar
Similar
Similar
Similar
Similar
Similar
arcs
437
93
285
426
287, 564
1002
288
27
1015
31
28
31
406
565
31
31
289
290
401, 405
38
84
565
877
.
of revolution
cylinders of revolution
cones
.
.
863
419
.811, 1022
sectors
565
Similar segments
565
Similarity of triangles
.431
Similitude,ratio of
418, 1023
Slant height,of frustum
of cone
881
of frustum
of pyramid
765
of pyramid
764
of right circular cone
865
Small circle of sphere
905
polygons
polyhedrons
.
.
.
.
.
.
.
.
.
.
.
.
.
.
....
"Solid,geometric
$olid angle
.
.
.
.
2, 9,
Art.
Solid
geometry
Solution, of exercises
of locus problems
of problems
602
275
143
123
79
900
....
....
of theorems
Sphere
of
circumscribed
center
diameter
901
about
hedron
poly929
901
904
927
of
great circle of
inscribed
in polyhedron
line tangent to
plane tangent
to
.
.
921
921
....
radius of
small circle of
surface of
volume
of
901
905
970
996,c
Spheres, tangent externally
tangent internally
.
.
....
tangent
to
each
other
.
.
.
Spherical angle
Spherical cone
Spherical degree
Spherical excess
Spherical polygon
angles of
diagonal of
936
936
937
sides of
936
vertices of
936
Spherical polygons, symmetrical 956
1010
Spherical pyramid
base of
1010
of
vertex
1010
1002
Spherical sector
base of
1002
1015
Spherical segment
altitude of
1016
bases of
1015
of one
base
1016
970
Spherical surface
938
Spherical triangle
952
birectangular
953
trirectangular
1006
Spherical wedge
1006
angle of
base of
1006
225
Square
569
Squaring the circle
69
Straight angle
Straight line
21, 23
determined
25
630
oblique to plane
629
parallel to plane
619
perpendicular to plane
21
Straightedge
of divisions
of circumference
Summary,
....
.
.
.
.
....
.
.
11
692
922
922
922
917
1003
983
989
of equal triangle theorems
.
529
118
INDEX
488
Art.
943
96
93
938
94, 99
99
93-98
431
948
732
749, 750
Akt.
Summary,
Triangle, polar
of plane
of formulas
570
geometry
197
parallel line theorems
243
quadrilaterals
431
similar triangle theorems
Exs. 320, 321
trapezoids
174
unequal angle theorems
175
unequal line theorems
106
103,
19,
Superposition
948
triangles
Supplemental
69
angles
Supplementary
72
Supplementary-adjacent angles
3, 8, 11
Surface
of
of
of
of
of
of
.
.
....
.
.
.
.
....
.
.
.
....
.
714
837
closed
conica
curved
35
819
970
cylindrical
of sphere
32, 33, 34
plane
prismatic
pyramidal
724
744
32
707
Surfaces
polyhedral angles
spherical polygons
common
....
common
....
length of
angle
Tetrahedral
Tetrahedron
Theorem
of
opposite of
solution of
converse
.
.
.
.
.
...
.
.
.
.
...
.
.
.
altitude of
bases of
956
of
Triangle
acute
altitudes of
bisectors of angles of
of
centroid
construction
determined
of
....
equilateral
isosceles
median
center
of
medians
of
notation
obtuse
parts of
79
386
493
181
222
.221
229
229
.
.
227
isosceles
median
.
136
Third proportional
Transformation
Transversal
Trapezium
Trapezoid
spherical
angle of
-
.
vertex
vertex
....
of
of
....
.
Triangles, classified
similarity of
supplemental
Triangular prism
Triangular pyramid
Trihedral
angle
....
.
.
.
698
699
700
699
699
birectangular
isosceles
rectangular
trirectangular
Trirectangular spherical triangle 953
736
Truncated
prism
753
Truncated
pyramid
....
Unit, of
measure
335, 466
....
769
of volume
.
286
333
333
319
698
719, 750
57
135
Tangent
external
internal
scalene
.
.
Symmetrical
Symmetrical
right
251
92
98
100
.126
265
269
133, 269
95
94
265
102
270
97
.
269
Variable
.
.
346, 348, 585-594
.
approaches its limit
independent
350
346
349
351
352, 594
38
840
838
limit of
reaches
its limit
related
Vertex, of angle
of cone
of conical surface
....
polyhedral angle
pyramid
pyramidal surface
spherical pyramid
triangle
Vertex angle of triangle
Vertical angles
Vertical polyhedral angles
693
of
of
of
of
of
Vertices, of
.
.
.
.
.
.
748
745
.
.
.
.
.
.
.
1010
99
94, 99
70_
.
.
lune
708
983
84
717
of polygon
ron
of polyhe
936
of spherical polygon
b
892,
of
cone
Volume,
.888
of cylinder
774
of rectangular parallelopiped
770
769,
of solid
.
.
.
of sphere
unit of
.
.
.
.
.
996, c
769
.
Wedge,
spherical
Zone
altitude of
of
bast
base
of one
1006
975
976
975
977
STANDARD.
MILNE'S
ALGEBRA
By
J.
WILLIAM
New
the
State
York
Ph.D.,
MILNE,
Normal
LL.D.,
President
N.
College, Albany,
of
Y.
$I.OO
Standard
THE
of
courses
is
Algebra
The
study.
followed, but
conforms
inductive
declarative
to
method
the
most
of
and
statements
recent
presentation
observations
ing
this kind of unfoldto
questions. Added
and
development of the subject are illustrative problems
and
points,the whole being
explanationsto bring out specific
driven
home
practice.
by varied and abundant
tional
^f The problems are fresh in character, and besides the tradidrawn
from
a
physics,
large number
problems include
ing
life.
and commercial
They are classified accordgeometry,
of the equations involved, not
the nature
to
according to
are
used, instead
The
subject matter.
principlesis
except
of
clear
some
pupil.
^[ Accuracy
and
of necessary
statement
concise, but
important ones,
are
the
definitions
proofs
left for the
of
maturer
and
of
principles,
years
of
the
encouraged by the use of
checks
and
sults
numerous
tests, and
by the requirement that reThe
be verified.
subjectof graphs is treated after simple
of their simple uses
in repreby some
equations,introduced
senting
in
related
in
a
nd
statistics,
picturing two
quantities
of change, and again after quadratics. Later they
the process
utilized in discussingthe values
of quadratic
are
expressions.
Not
Factoring receives
particularattention.
only are the
usual cases
given fullyand completely with plenty of practice,
but
the
self-reliance
theorem
is
are
taught.
made
of pointed
helpfuland frequent reviews
are
up
oral questions,
abstract exercises, problems, and recent
college
examination
entrance
questions. The book is unusuallyhandy
in size and convenient
for the pocket. The
page size is small.
^f
The
factor
and
ROBBINS'S
PLANE
TRIGONOMETRY
EDWARD
By
R.
William
Senior
ROBBINS,
Perm
Charter
School,
Mathematical
ter,
Mas-
Pa.
Philadelphia,
$0.6o
book
is intended for
THIS
thorough familiarity
with
beginners.
the
essential
skill in operating
with
satisfactory
is illustrated in the usual
than
clear-cut
usually
The
^[
and
it is
work
concise
and
is sound
but
those
to
givea
truths, and
It
processes.
diagrams are
the
a
more
elucidating.
teachable,and is written in clear
and
understood.
stylethat makes it easily
has been proved,
Immediatelyafter each principle
pressed
appliedfirstin illustrativeexamples,and then further imexercises.
ment
Accuracy and rigorof treatby numerous
language,in
shown
are
matter
to
maimer,
It aims
in every detail,and all irrelevant and extraneous
is excluded, thus giving greater prominence
universal rules and
"fjThe
a
formulas.
Geometry precedingthe first
of
chapter are invaluable. A knowledge of the principles
taken
is,as a rule,too freely
geometry needed in trigonometry
for granted. The
author givesat the beginningof the book
of the appliedprinciples,
with reference to the
a statement
sections of his Geometry, where
such theorems
are
provedin
full. Cross
references in the text of the Trigonometryto
make
those theorems
it easy for the pupilto review
to
or
knowledge.
supplementimperfect
^| Due emphasisis given to the theoretical as well as to the
of exnumber
of the science.
The
amples,
practical
applications
references
both
as
many
AMERICAN
(313)
next
abstract,is far in
the market.
on
exercises
having the
and
concrete
in other books
Plane
to
as
most
lowest
This
books, and
book
excess
contains four times
twice
as
many
number.
BOOK
of those
COMPANY
as
that