THEOREMS
IRT ANGLE
ITE OREM
6.9
If
line
a
:
-
-
drawn
is
other two
sides
parallel
to
divided
are
side
one
in
of
the
a
triangle
ratio
same
the
to intersect
other two sides
in
distinct
points the
,
A
.
Ng
M
'
.
-
-
-
D f
.
'
-
.
-
-
.
.
×
-
.
-
t
B
-
-
-
-
-
-
I E
.
"
-
-
-
-
-
-
y
'
y
*
Ioof
We
D
-
-
-
'
'
,
.
.
join
us
,
area
triangle
of
prove
BE
A ADE
ABC
which
in
that
and
IBI LIFE
CD
f- Iz
=
base
ar
,
parallel
to
x
draw DM I AC and
height
)
-
-
side
.
intersects
BC
DBXEH
I
A- Ex DM
=
-
-
ARCADES
ar
(BDE)
=
IAD
x
and
( DEC )
EN
EN t AB
.
A- DX EN
( BD E) I
ARCADE)
Therefore
line
.
then
and
=
,
a
.
ARCADE ) I
So ,
Similarly
a
respectively
need to
Let
.
i.
C
E
and
Now
.
given
are
We
-
:
D×EN
g
-2 DB
×
EN
ar
=
AI
DB
-
-
IEC
x
DM
.
other two sides
AB
and AC
at
and
arCADEI=IAExD
,
Note that
DBDE and DEC
/
theorem 6.1
If
line
a
third
side
IEC
( DEC )
ar
x
the
divides
two sides
any
of
If
-
(
or
can
-
-
-
-
-
-
-
-
-
-
-
-
-
same
ratio
then the
,
line
is
parallel
to
the
'
-
-7
E
y
to
proved by taking
be
BC
a
line
DE such
that
parallel
to BC
and
A¥=f¥Ez
assuming
that
.
BC , draw
to
parallel
Aage
,
=
a
line
DE
'
.
AI
'
E- C
Aet =Ae
,
in two
and DE
c
DE is not
-
BC
parallels
same
A
parallel
THEOREM
the
.
This theorem
Therefore
in the
triangle
a
B
So
between
.
.
If
and
DE
o
↳P
not
base
same
o
E
is
AI
EC
DM
on
are
=
6.38
-
triangles corresponding angles
proportion)
,
and hence
the
two
are
equal
triangle
,
are
then
their
similar
corresponding
sides
in the
are
same
D
A
.
P•
.
B
C
E
-
-
-
-
-
-
-
-
•
-
Q
F
ratio
DE
.
Given : Two
C
to prove
triangles
Construction :
In
Draw
P and
and
DDPQ
AB
-
LA
DABC I
LB
=
Hence
Q
LP
,
PO
ioin
-
D
B
,
-
-
E
and
such that
DP
-
-
AB
and
DA
-
-
.
LD
D
DQ
A
LP
-
*
=
C
-
-
-
-
-
-
-
-
•
-
9
F
E
LE
PA
and EF
with transversal
PE
PQHEF
Theorem
-
)
)
B
,
DE and DF
on
and
DDPQ
LB LE
For lines
By
-
-
Thus
-
DP
-
=
AC
,
A
:
DABC
But
that
such
DDEF
-
respectively
Proof
DDEF
and
F
=
A ABC
:
DABC
6.9
*
FEI fff 7¥ FI
-
-
-
Adding
-
,
l
'
both the
on
P +1=7%+9
PEypD#
DI
DP
=
=
QftDQ
DO
DI
DQ
sides
:
,
LP and LE
are
corresponding angles
,
and
they
are
equal
.
AC
DI
DI
=
DF
DE
And
DP
construction
by
AI
=
AB
and
DQ
-
-
AC
AI
=
DF
DE
Similarly
can
we
,
Are
that ;
BI
=
DE
Therefore
prove
Ef
,
¥I=¥÷=B¥
Since
o
,
all
sides
~
DDEF
D ABC
:
Hence
proved
THEOREM
angles
are
:
-
triangles
in two
If
:
.
6.4 (Sss )
-
-
proportion
in
are
equal
sides
,
of
one
triangle
the two
and hence
D
proportional
are
triangle
to
similar
are
the
sides
of
the
other
AC
respectively
triangle
,
then their
corresponding
.
A
P
B
•-
-
-
-
-
-
-
•
- -
Q
F
C E
Given :
A ABC
To
prove
:
LD
L B
,
such
that
¥I= Iffy ,f¥
=
-
-
LA
DDEF
and
=
,
-
-
Construction
Draw
P
LE
,
LC
-
LF
and D ABC
~
A DEF
.
0
o
and
Q
on
DE
and DF
such
that
DP
-
-
AB
and
DO
-
and
join
PA
.
Prod :
Given ;
AI
CI
=
DF
DE
And
DP
DI
AB
-
-
-
-
AC
DI
=
DE
DE
DI
DO
,
=
DI
DP
DQ
Subtracting
7¥
9
a
-
DE
-
both
on
-
-
FI
DP
=
DQ
QI
DO
DP
DP
PI
Using
DO
-
-
DP
PI
a
-
DF
=
-
sides
=
DO
FF
Theorem
6.2
PQHEF
=
Now
.
,
PQ
For lines
and
with transversal
LP
In
=
LE
DDPQ and
LP
=
-
EF
,
PE
④
D. DEF
LE
L D= LF
DDPQ
"
n
DDEF
fed '=I÷
'
=
,
for
lines
PA and EF
with transversal
LA
-
LF
OF
-
,
Also
,
¥÷=¥f=f¥
AB
and
thus
DP
-
-
}
and CA
-
-
Given
QD
¥7 Eee Eff
-
-
,
From
-
-
and
7¥
EF
BC
:
PI
=
Pcs
=
In DABC
-
AC
BC
DABCE
.°
.
LB
=
L
=
C
LA
DDPO ;
and
AB
DP
=
DO
=
=
④
pg
DDPQ
LP
LO
LD
-
But from ④
LP
Therefore
,
LB
LC
and
Therefore
In
D ABC
LB
LC
i
.
DABC
Hence
LQ= LF
and
LE
=
-
-
=
LP
Lo
=
-
,
and
=
LE
=
LF
n
A DEF
D DEF
proved
.
CE
L F
-
①
THEOREM
6.5 ( SAS )
-
-
If
angle of
one
proportional
8
-
triangle is
triangles
a
the
then
equal
are
to
similar
angle of
one
the
other D
and
sides
including
these
angles
are
.
D
✓
A
U
P
.
B
C
Given
Two
,
DABC
-
-
-
-
- -
Q
a
-
F
and
DDEF
Draw
:
such
~
P and Q
,
=
CI
DF
DE
DP
Dd
=
je
DF
7¥
-
-
DI
DB
Subtracting
7¥ t.FI
9
on
-
DE
-
DP
DP
-
-
=
i
DQ
AI
DQ
pet
-
the sides
=Dt
Dpt
PE
both
Dose
OF
that
DDEF
:
AI
And
-
A ABC
Construction
Given
-
E
8
prior
Pref
-
:
triangles
To
.
LA
-
LD
,
DAEI
=
Appt
.
on
DE
and DF
such that
DP AB
-
and
DO
=
Ac
respectively
and
join
Pa
.
Using
Theorem
6.2
PO
=
HEE
Now
.
,
For lines PA
with
LP
Now
LE
DDPQ
and
AB
LB
:O
DDPQ
-
-
LQ
from ⑨
LP =
LE and
Therefore
-
LF
=
LP
-
LE
-
LC =L 9=2 F
and
Therefore
-
⑤
①
,
In
DABC
LB
LC
D ABC
Hence
LO
,
LB
:
DO
=
LP
=
(C
LD
=
AC
DABC ±
DP
=
LA
o
⑨
-
,
In DABC
But
PE
transversal
=
)
EF
and
E
LE
=
-
LF
DDEF
proved
and DDEF
.
( from iv )
(
from v )
For lines
PO
and
with
transversal
LO
LE
=
-
EF
OF
THEOREM
6.6
:
-
-
The
ratio
the
of
P
'
E pree
i.
:
i
i
im
D
ABC
-
in
Q
c
µ,
:
equal
to
the
square
I
.
Given
is
!
A
B
triangles
similar
two
of
areas
R
N
D POR
aarrff.BA#=fApBaY=fBacj=fAfj
:
Construction
Draw
:
AM 1. BC
and
PNIQR
-
Ioof :
ar
( ABC)
I
×
Base
=L
x
BC
=
①
Dividing
×
x
Height
AM
and
ar
# Bc)
I
X
( POR)
Iz
×
(POR)
⑨
-
BC
x
I
x
Base
=L
×
OR
=
AM
ar
=
-
ar
arCAB#
ar
In D ABM
LB
D ABM
o
:
Afif
=
=BCxA
( POR)
AR
-
PN
X
and DPAN
LO
-
N
QRX PN
,
LM
-
-
LN
D PON
AI
BO
-
PN
From CA)
arCAB#
arcpgpg
From
=
BCx
Pax
PN
(B)
aarrff.fr#-- :#
¥
x'
.
-
②
AO
x
x
Height
PN
-
④
of
ratio
of
their
corresponding
sides
.
Now
,
Given
DABC
a÷= :#
Putting
D POR
N
'¥
=
Ccl
aaieap.fi#axtEa--fA*
Now
Aptos iff 7¥
again using
,
'
=
-
-
adf.is#=faH=EsE=GAaT
toured
Hence
-
.
6. F
THEOREM
o
o
-
-
If
on
perpendicular
a
sides
both
is
from
perpendicular
the
of
drawn
the vertex
are
similar
of
the
to the
right angle of
whole
triangle
a
right
and
to
D
to
the
each other
hypotenuse
.
B
:
"
÷
A
Given
DABC
8
to pride
right angled
DAD B
8
n
DBDC
and
Ioof
In
LA
n
D ABC
D BDC
and D ABC
LA
=
LA DB
D ADB
D ADB
N
DA BC
:
DADB
~
=
L ABC
DABC
C
D
-
④
at
B
and
perpendicular from
B
intersecting
AC
at D
.
then
right
D
Similarly
.
DBDC
In
D ABC
and
L C
( BDC
third
n
Hence
a
DABC
~
triangle
another
the
,
second
triangle
is
similar
the
to
Therein )
Pythagoras
square of the
D. the
hypotenuse
is
equal
the
to
sum
the
of
squares
l
l
l
l
'
7
A ABC
:
D
.
AB't Be
Draw
8
at B
right angle
BDI AC
:
Since
Using
ACE
8
C
'
A
Perot
then
:
l
Construction
,
.
Y
to prove
D
.
B
Given
third
DBDC
6£ (
right
to
similar
are
proved
theorem
If
-
DBDC
and
is similar
triangle
DAD B
%
DA BC
N
D
one
D ABC
~
and
DAD B
If
LA BC
=
D BDC
From ①
LC
=
BDI
Theorem
.
AC
6.7
DADB
since
D
~
D ABC
the sides
,
are
in
AI
the
=
AB
AD
.
AC
of
DBDC
similar
same
ratio ,
since
are
.
~
the
in the
Are
CD
AC
BC
=
A.BZ
-
⑨
DABC
CD
.
sides
same
=
of
ratio
similar
,
Be
AC
AC
=
BCZ
-
D
of
other
two
sides
first
and
Theo REM 6
9 8
.
-
-
In
D
a
if
,
side
one
side
first
to the
opposite
of
square
is
a
is
equal
to
right angle
the
sum
of
the
square
of
the
other two
sides
,
then the
.
A
B
Given
C
A
8
II prove
triangle
LB
%
Construction
=
ABC in which
BC
'
900
Draw
8
ABI
ACE
D POR
right angled
at
Q
,
such
A
B
Pref
:
In DPQR
,
29=900
By
Theorem
Pythagoras
PRZ
Since
PO
P R2
Also
that
AC
From ④
=
and
PRZ
PR
BR
-
ABI Ba
=
given
and
AB
-
-
,
Pg 't BRZ
=
=
=
BC
-
.
AB't BC
"
④
ACZ
AC
-
-
④
that
PA
-
-
AB
p
C
g
R
and
OR
=
Bc
.
angle
In
A ABC and D PAR
AC
AB
BC
=
-
-
=
PR
Pg
OR
D ABC E D POR
:
o
LQ
L Be
Since
:
.
LB
Hence
20=900
=
900
proved
.