Chapter 5
Chapter 5 Homework
5.1
2
5
6
12 points
20 points
10 points
5.2
3
10 points
5.3
no problems assigned
5.4
2
4
10 points
25 points
5.5
7
14 points
5.6
2
6
8
08 points
10 points
16 points
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We will keep the 6 axioms of Neutral Geometry and add a 7th: The Euclidean Parallel Postulate.
Now we are in Euclidean Geometry. These 7 axioms are the background assumptions for this
whole chapter. In Chapter 6 we will CHANGE the 7th axiom to the Hyperbolic Parallel
Postulate.
5.1
Basic Theorems of Euclidean Geometry
Theorem 5.1.1
If two parallel lines are cut by a transversal, then both pairs of alternate
interior angles are congruent.
Theorem 5.1.2
If l and l’ are two lines cut by a transversal t such that the sum of the
measures of the two interior angles on one side of 6t is less than 180  ,
then l and l’ intersect on that side of t.
Theorem 5.1.3
For every  ABC,  (ABC )  180  .
Theorem 5.1.4
If  ABC is a triangle and DE is any segment, then there exists a point F
such that  ABC  DEF .
Theorem 5.1.5
If l and l’ parallel lines and t  l is a line such that t intersects l, then t also
intersects l’.
Theorem 5.1.6
If l and l’ parallel lines and t is a transversal such that t  l , then t  l '.
Theorem 5.1.7
If l, m, n, and k are lines such that k l , m  k , and n  l , then either
m  n , or m n .
Theorem 5.18
If l m and m n , then either l  n or l n.
Theorem 5.1.9
There exists a rectangle.
Note that most of these are logically equivalent to the Euclidean Parallel Postulate. The
following theorem is the first new result from making our choice of the 7th axiom.
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Theorem 5.1.10
Properties of Euclidean Parallelograms
If ABCD is a parallelogram, then
1. The diagonals divide the quadrilateral into two congruent triangles
( ABC  CDA and ABD  CDB ).
2. The opposite sides are congruent.
3. The opposite angles are congruent.
4. The diagonals bisect each other.
D
C
A
B
Note that the Euclidean Saccheri Quadrilateral is a rectangle and has all of these properties.
All SQ’s are parallelograms, but not all parallelograms are SQ’s.
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5.2
The Parallel Projection Theorem
Theorem 5.2.1
Let l, m, and n be distinct parallel lines. Let t be a transversal that cuts
these lines at point A, B, and C respectively and let t’ be a transversal that
cuts the lines at A’, B’, and C’ respectively. Assume A  B  C , then
AB A ' B '

AC A ' C '
This is setting the stage for similar triangles.
A
B
A'
B'
C
C'
When you project with parallel lines you get proportional changes.
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Lemma 5.2.2
Let l, m, and n be distinct parallel lines. Let t be a transversal that cuts
these lines at point A, B, and C respectively and let t’ be a transversal that
cuts the lines at A’, B’, and C’ respectively. Assume A*B*C. If
AB  BC , then A ' B '  B ' C ' .
A
B
C
A'
B'
C'
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Here’s a typical problem:
12
5
x
x+6
Solve for x!
Note the two transversals…proportion problems bringing algebra into geometry.
Note that in our text there are two cases, the rational case and the irrational case. That’s very
interesting and the proofs are not hard to comprehend. I recommend them to your attention.
COURSEWORK #1, Chapter 5.
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5.3
Theorem 5.3.1
Similar Triangles
If  ABC and DEF are two triangles such that  ABC ~ DEF , then
AB DE

AC DF
This is built on parallel projection!
C
D
A
Corollary 5.3.2
B
If  ABC and DEF are two triangles such that  ABC ~ DEF , then
there is a positive number r such that
DE  r ( AB), DF  r ( AC ), and EF  r ( BC ).
R is called the constant of proportionality.
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Theorem 5.3.3
SAS Similarity Criterion
If  ABC and DEF are two triangles such that CAB  DEF and
AB / AC  DE / DF , then  ABC ~ DEF .
Theorem 5.3.4
Converse to Similar Triangles Theorem
If  ABC and DEF are two triangles such that AB / AC  DE / DF , then
 ABC ~ DEF .
Additional helpful theorem
AA similarity:
If two pairs of corresponding angles angles are congruent, then the
triangles are congruent!
Why is this a Euclidean only criterion?
CourseWork Chapter 5 #2
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5.4
The Pythagorean Theorem
Theorem 5.4.1
If  ABC is a right triangle with a right angle at vertex C, then
a 2  b2  c2 .
Three similar right triangles are created by the altitude of any right triangle.
We will use this construction at the end of this section in a proof of the
Pythagorean Theorem.
D
B
A
mADB = 90.00
C
D
ΔACD ~ ΔABD
They share angle A and both have one
right angle. AA Similarity.
B
A
mADB = 90.00
C
ΔCBD ~ ΔDBA
D
They share angle B and both have one
right angle. AA Similarity
B
A
mADB = 90.00
C
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ΔACD ~ ΔDCB
Which is to say, the two smaller triangles
are also similar.
D
Note that  A and  B are complements.
 A = 90° −  B.
In ΔACD, we have a right angle,  A and
 ADC = 90° −  A =  B.
B
A
C
mADB = 90.00
In ΔDCB, we have a right angle and  B.
So, by the AA Similarity Theorem, these are similar.
Thus, by constructing an interior altitude, you get 3 pairs of similar right triangles.
Also note that the following ratios hold:
D
B
A
C
mADB = 90.00
AB BD

BD CB
hypotenuse
shortest leg
AB AD

AD AC
hypotenuse
medium leg
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Pythagorean Theorem
In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the
length of the hypotenuse.
C
ABC is a right triangle with altitude CD
a
b
x
c-x
A
D
B
c
We know from previous work that the three triangles are similar and that
c a

a x
c
b

b cx
we are using the ratios:
hypotenuse
hypotenuse
and
medium leg
shortest leg
Working with the equalities we have
a 2  cx
b 2  c 2  cx
a 2  b 2  cx  c 2  cx  c 2
This proves the theorem.
In 1947, Elisha Scott Loomis published a book with over 200 proofs of the Pythagorean
Theorem. NCTM recently put it back in print.
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Theorem 5.4.3
The height of a right triangle is the geometric mean of the lengths of the
projection of the legs.
“height” is a special case of a cevian line, the altitude.
The geometric mean of two numbers x and y is
xy . What are the “projections of the legs?
How would you show that h  xy ? CourseWork #3
C
a
b
x
A
y
D
c
B
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Theorem 5.4.4
The length of one leg of a right triangle is the geometric mean of the
length of the hypotenuse and the length of the projection of that leg onto
the hypotenuse.
Theorem 5.4.5
If  ABC is a triangle with a 2  b 2  c 2 , then  ABC is a right triangle.
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5.5 Trigonometry
Theorem 5.5.2
Pythagorean Identity
For any angle  , sin 2   cos 2   1 .
Theorem 5.5.3
Law of Sines
sin A sin B sin C


a
b
c
Recall the ambiguous case! This can be another version of SSA! BEWARE!
Theorem 5.5.4
Law of Cosines
If  ABC is any triangle, then c 2  a 2  b 2  2ab cos C.
Let’s look at WHY this is a generalization of the Pythagorean Theorem
CourseWork #4
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5.6
Exploring the Euclidean Geometry of the Triangle
Definition:
Concurrent lines intersect at one interior point. We are generally speaking
of 3 lines, not 2.
Theorem 5.6.2
Median Concurrence Theorem
The three medians of any triangle are concurrent; that is, if  ABC is any
triangle and D, E, and F are the midpoints of the sides opposite A, B, and
C, respectively, then AD, BE , and CF all intersect in a common point
G. Moreover, AG  2GD, Bg  2GE , and CG  2GF .
A median for a triangle is a segment joining a vertex and the midpoint of the line opposite that
vertex. The point of concurrency of the medians is the centroid. The
centroid is always a point of the interior of the triangle.
The smaller interior triangle formed by joining the median points is called
the medial triangle.
An altitude for a triangle is a segment through a vertex and perpendicular to the side opposite
that vertex. The point of concurrency for the altitudes is the orthocenter.
The orthocenter can be inside, on, or exterior to its triangle. The triangle
formed by joining the feet of the perpendiculars is called the orthic
triangle. The orthic triangle is not necessarily contained in the original
triangle.
A perpendicular bisector is a line that is through the midpoint of a side and perpendicular to
that side. The point of concurrency of the perpendicular bisectors is called
the circumcenter. The circumcenter may be interior, exterior or on the
triangle.
Orthocenter
Centroid
Circumcenter
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mDEF = 89.66°
mDEF = 61.13°
E
Orthocenter
Centroid
Circumcenter
D
F
mDEF = 61.13°
E
Circumcenter
Orthocenter
Centroid
D
F
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Theorem 5.6.3 Euler Line Theorem
The orthocenter H, the circumcenter O, and the centroid G of any triangle
are collinear. Furthermore H*G*O (unless the triangle is equilateral in
which case the three points coincide) and HG = 2GO.
These three are not the only “centers” for a triangle. There are many others.
Chapter 5 CourseWork #5 Come up with a mnemonic way to remember the four centers!
Definition:
Cevian line: a line segment through a vertex and intersecting the side
opposite that vertex. A Cevian line is proper if it only passes through one
vertex. Cevians are named by the vertex passed through and the point of
intersection with the side opposite. An altitude is a cevian line.
Theorem 5.6.4
Ceva’s Theorem
Let  ABC be any triangle. The proper Cevian lines
AL, BM , and CN are concurrent or mutually parallel if and only if
AN BL CM


 1.
NB LC MA
Cevian triangle:
Draw triangle ABC. Extend all 3 sides to be inifinite lines. Pick a point P that is not on any of
these lines. Emanating from P through the 3 vertices in turn connect P to
the side opposite the vertex. (e.g. PA will intersect BC at point L).
C
L
P
A
B
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Do this for M and N. The resulting triangle is called the Cevian triangle.
C
N
L
P
A
M
AN = 5.14 cm
BL = 5.04 cm
NB = 3.48 cm
LC = 1.83 cm
AN
NB
= 1.48
CM = 1.36 cm
AM = 5.56 cm
CM
= 0.25
AM
BL
LC
B
= 2.76
AN BL CM
∙
∙
= 1.00
NB LC AM
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AN = 10.32 cm
BL = 3.84 cm
NB = 3.31 cm
LC = 1.74 cm
AN
NB
= 3.12
CM = 0.96 cm
AM = 6.59 cm
CM
= 0.15
AM
BL
LC
= 2.21
AN BL CM
= 1.00
∙
∙
NB LC AM
M
C
L
P
A
B
N
Either 0 or exactly 2 points of the Cevian triangle can be outside the original triangle when P is
moved outside the triangle. If N is on the outside of triangle ABC then the ration AN/NB we
say that we “sense” the ratio as a negative and we report it with a negative sign as a signal to the
reader that N is on the extended ray AB . (This is for A*B*N) If A*N*B the sensed ratio is
positive. If you have one ratio sensed as negative, then you’ll have a second one, too, so that
the product of the ratios is positive 1.
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Menelaus Points
C
M
N
L
A
B
Here we have 3 points that are collinear, rather than concurrent lines. If you extend the sides of
the triangle to become lines, you can run one line through the 3 extended sides picking up 3
collinear points. These are called Menelaus Points. Proper Menelaus Points are not equal to any
vertex of the triangle.
Note that A*B*L is sensed negative!
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Theorem 5.6.5
AN = 13.68 cm
BN = 2.91 cm
BL = 3.21 cm
LC = 6.00 cm
AN
BN
BL
LC
Theorem of Menelaus
Let  ABC be any triangle. Three proper Menelaus points L, M, and N
on the lines BC , AC , and AB are collinear if and only if
AN BL CM


 1.
NB LC MA
= 4.70
C
= 0.54
M
CM = 2.02 cm
MA = 5.08 cm
CM
MA
= 0.40
L
N
A
B
AN BL CM
∙
∙
= 1.00
BN LC MA
Sketchpad isn’t programmed to sense negative or positive so we have to
add in the negative sign ourselves!
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Theorem 5.6.6
Let  ABC be any triangle, then the associated Morley triangle is
equilateral.
The Morley triangle is an amazing construction. Note that he figured it all out without anything
but a pencil and a ruler and logic!
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