Module 6

Module 6 - SIMILARITY
Regional Mass Training of Grade 9 Mathematics
Teachers
May 17 – 21, 2014

1. Solve problems involving similar polygons.
2. Prove certain triangles are similar by using AA,
SSS, and SAS.
3. Solve problems involving Basic Triangle
Proportionality Theorem and Triangle Angle
Bisector Theorem.
4. Solve problems involving Right Triangle
Similarities and Special Right Triangles.
DEPARTMENT OF EDUCATION

Warm Up
Solve each proportion.
1. 2. 3.
4. If ∆QRS ~ ∆XYZ, identify the pairs of congruent angles and
write 3 proportions using pairs of corresponding sides.
Review: Proportion
z = ±10 x = 8
Q  X; R  Y; S  Z;

Similar Polygons
Two polygons are said to be similar if and only if:
i. corresponding angles are congruent.
ii. corresponding sides are proportional.
A
B
C
D
E
M
N
O
P
Q ∠A↔∠M
∠E↔∠Q
∠C↔∠O
∠D↔∠P
∠B↔∠N
AB↔ MN
BC↔ NO
CD↔ OP
DE↔ PQ
AE↔ MQ
∠A≅ ∠M
∠E≅ ∠Q
∠C≅ ∠O
∠D≅ ∠P
∠B≅ ∠N
AB
MN
=
BC
NO
=
CD
OP
=
DE
PQ
=
AE
MQ
ABCDE∼MNOPQ

Similar Polygons
Example: If ABCDE∼MNOPQ, determine the value of x and
y. Figure is drawn not to scale.
A
B
C
D
E
Answer:
y=3
x=2
9
M
N
O
P
Q
6
y+3
4
x+2
5
2x+2 6
7.5
4y-3

Triangle Similarities

Example 1: Using the AA Similarity Postulate
Explain why the triangles are
similar and write a similarity
statement.
Since 𝐴𝐵 ∥ 𝐷𝐸, B  E by the Alternate Interior
Angles Theorem. Also, A  D by the Right Angle
Congruence Theorem. Therefore ∆ABC ~ ∆DEC by AA~.
Similar Triangles

Similar Triangles
In ∆ABC and ∆DEF,
AB
DE
=
BC
EF
=
AC
DF
.
Prove that ∆ABC∼∆DEF.
B
A
C
D
F
E

Similar Triangles
Proof:
Construct GH in ∆DEF
such that G and H are in
DE and EF respectively,
GH ∥ DF, GE ≅ AB and
BC ≅ EH.
B
A
C
D
F
E
G H

Similar Triangles
B
A
C
D
F
E
G H
By AA Sim. Postulate,
∆EGH∼∆EDF which
implies
EG
ED
=
GH
DF
=
EH
EF
.
Since EG ≅ AB, BC ≅ EH,
then
EG
ED
=
AB
ED
and
EH
EF
=
BC
EF
by substitution.

Similar Triangles
B
A
C
D
F
E
G H
And since
AB
ED
=
AC
DF
from
the given and
EG
ED
=
GH
DF
,
then by transitivity PE,
AC
DF
=
GH
DF
which implies
AC=GH by multiplication
PE.

Similar Triangles
B
A
C
D
F
E
G H
Meaning ∆ABC ≅∆GEH
by SSS congruence
postulate. Since
∆EGH∼∆EGF, then
∆ABC∼∆DEF by
substitution.

Similar Triangles
Given ∆ABC and
∆DEF such that
AB
DE
=
AC
DF
and ∠𝐴 ≅ ∠𝐷
B
A
C
D
F
E

Similar Triangles
B
A
C
D
F
E Locate P on DE so that
PE=AB. Draw PQ so that
PQ║ DF. By the AA Sim.
Theo., we have
∆PEQ∼∆DEF, and thus
DE
PE
=
EF
EQ
=
FD
QP
.
P Q

Similar Triangles
B
A
C
D
F
E
P Q
Since PE=AB, we can
substitute this in the
given proportion and
find EQ=BC and QP=CA.
By SSS congruence
Theo., it follows that
∆PEQ≅∆ABC.

Similar Triangles
B
A
C
D
F
E
P Q
And since ∆PEQ∼∆DEF
and ∆PEQ≅∆ABC, by
substitution, then
∆ABC∼∆DEF

Example : Verifying Triangle Similarity
Verify that the triangles are similar.
∆PQR and ∆STU
Therefore ∆PQR ~ ∆STU by SSS ~.
Similar Triangles

Example : Verifying Triangle Similarity
∆DEF and ∆HJK
Verify that the triangles are similar.
D  H by the Definition of Congruent Angles.
Therefore ∆DEF ~ ∆HJK by SAS ~.
Similar Triangles

A  A by Reflexive Property of , and B  C
since they are both right angles.
Example : Finding Lengths in Similar Triangles
Explain why ∆ABE ~ ∆ACD, and
then find CD.
Step 1 Prove triangles are similar.
Therefore ∆ABE ~ ∆ACD by AA ~.
Similar Triangles

Example Continued
Step 2 Find CD.
Corr. sides are proportional.
Seg. Add. Postulate.
Substitute x for CD, 5 for BE,
3 for CB, and 9 for BA.
Multiplication PEx(9) = 5(3 + 9)
Simplify.9x = 60
Similar Triangles

Example : Writing Proofs with Similar Triangles
Given: ∆BAC with medians
𝑫𝑪 and 𝑨𝑬.
Prove: ∆BAC ~ ∆BDE
Similar Triangles
A
B
C
D E

Example Continued
Similar Triangles
Statement Reason
1. 𝐷𝐶 and 𝐴𝐸 are
medians of ∆BAC
1. Given
2. 2BD=BA ; 2BE = BC 2. Definition of
Medians
3.
𝐵𝐷
𝐵𝐴
=
1
2
;
𝐵𝐸
𝐵𝐶
=
1
2
3. Properties of Ratio
4.
𝐵𝐷
𝐵𝐴
=
𝐵𝐸
𝐵𝐶
4. Transitive PE
5. ∠𝐵 ≅ ∠𝐵 5. Reflective PE
6. ∆BAC ~ ∆BDE 6. SAS ~
A
B
C
D E

Check It Out!
Given: M is the midpoint of JK. N is the
midpoint of KL, and P is the midpoint of JL.
Similar Triangles

Statements Reasons
Check It Out! Example 4 Continued
1. Given
1. M is the mdpt. of 𝐽𝐾,
N is the mdpt. of 𝐾𝐿,
and P is the mdpt. of
𝐽𝐿.
2. Midline
Theorem2.
3. Mult.
PE.
3.
4. SSS ~
Step 34. ∆JKL ~ ∆NPM
Similar Triangles

Basic Triangle Proportionality
Proof:
In ∆ABC, let D and E be points on 𝐴𝐵 and 𝐵𝐶
respectively, such that 𝐷𝐸║ 𝐴𝐶. We have to prove that
𝐵𝐷
𝐷𝐴
=
𝐵𝐸
𝐸𝐶
. Since parallel lines form congruent
corresponding angles, we have ∠𝐷 ≅ ∠𝐴 and ∠𝐸 ≅ ∠𝐶.
By the AA Sim. Theo., we say ∆ABC∼∆DBE.

Since corresponding sides of similar triangles are
proportional,
𝐵𝐷
𝐵𝐴
=
𝐵𝐸
𝐵𝐶
. By Reciprocal Property, we have
𝐵𝐴
𝐵𝐷
=
𝐵𝐶
𝐵𝐸
. By Subtraction Property of Proportionality, we
have
𝐵𝐴−𝐵𝐷
𝐵𝐷
=
𝐵𝐶−𝐵𝐸
𝐵𝐸
and hence
𝐷𝐴
𝐵𝐷
=
𝐸𝐶
𝐵𝐸
or
𝐵𝐷
𝐷𝐴
=
𝐵𝐸
𝐸𝐶
.

Check It Out!
Find the value of x in the figure below. The figure is not
drawn to scale.
4
6
3
x
6
4
=
𝑥 − 3
3
4(x – 3)=18
4x – 12 =18
4x =30
x=
15
2

Triangle Angle Bisector Theorem
If a segment bisects an angle of a triangle, then it
divides the opposite side into segments proportional to
the other two sides.
Proof:
1 2
3
4
Let 𝐴𝑋 bisect ∠A of ∆ABC. We must
prove that
𝐵𝑋
𝑋𝐶
=
𝐴𝐵
𝐴𝐶
.
B
X C
A
Y
Draw 𝐵𝑌 parallel to 𝐴𝑋. Extend 𝐶𝐴 so
that it intersects 𝐵𝑌 at Y. Since 𝐵𝑌 ∥
𝐴𝑋, we have
𝐵𝑋
𝑋𝐶
=
𝐴𝑌
𝐴𝐶
. By theorems
involving parallel lines cut by a
transversal, we also have ∠1 ≅ ∠3
and ∠4 ≅ ∠2.

If a segment bisects an angle of a triangle, then it
divides the opposite side into segments proportional to
the other two sides.
Proof:
1 2
3
4
B
X C
A
Y
Since 𝐴𝑋 bisect ∠A, then ∠1 ≅ ∠2 and
∠4 ≅ ∠3. Thus, by the Isosceles
Triangle Theorem, AY = AB. By
substitution, we conclude that
𝐵𝑋
𝑋𝐶
=
𝐴𝐵
𝐴𝐶
.

Check It Out!
In the figure below, ∠𝐵𝐴𝑋 ≅ ∠𝐶𝐴𝑋. Use the lengths to
find the value of y.
𝐴𝐵
𝐴𝐶
=
𝐵𝑋
𝑋𝐶
14
y X C
9
A
15
15
9
=
𝑦
14 − 𝑦
15(14 – y) = 9y
y = 8.75
B

Consider ∆ABC such that ∠𝐶 is right and 𝐶𝐷 is
the altitude to the hypotenuse. We have to
prove that ∆ADC∼∆ACB∼∆CDB.
Since 𝐶𝐷 ⊥ 𝐴𝐵, then ∠𝐴𝐷𝐶 = ∠𝐶𝐷𝐵 = ∠𝐴𝐶𝐵 = 900.
Since ∠𝐴=∠𝐴, then by AA Sim. Theo., we have
∆ADC∼∆ACB. Using the same theorem and that
∠𝐵=∠𝐵, we have ∆ACB∼∆CDB. Hence,
∆ADC∼∆CDB by transitivity.
Right Triangle Similarity

Consider the proportion . In this case, the
means of the proportion are the same number, and
that number is the geometric mean of the extremes.
The geometric mean of two positive numbers is the
positive square root of their product. So the geometric
mean of a and b is the positive number x such
that , or x2 = ab.

You can use Theorem on Right Triangle Similarity to
write proportions comparing the side lengths of the
triangles formed by the altitude to the hypotenuse of
a right triangle. All the relationships in red involve
geometric means.

Example 3: Finding Side Lengths in Right Triangles
Find x, y, and z.
62 = (9)(x) 6 is the geometric mean
of 9 and x.
x = 4 Divide both sides by 9.
y2 = (4)(13) = 52 y is the geometric
mean of 4 and 13.
Find the positive square root.
z2 = (9)(13) = 117 z is the geometric
mean of 9 and 13.
Find the positive square root.

Once you’ve found the unknown side lengths,
you can use the Pythagorean Theorem to check
your answers.
Helpful Hint

Special Right Triangles

Example : Craft Application
Jana is cutting a square of material for a tablecloth.
The table’s diagonal is 36 inches. She wants the
diagonal of the tablecloth to be an extra 10 inches
so it will hang over the edges of the table. What
size square should Jana cut to make the tablecloth?
Round to the nearest inch.
Jana needs a 45°-45°-90° triangle with a hypotenuse
of 36 + 10 = 46 inches.

Check It Out! Example
What if...? Tessa’s other dog is wearing a square
bandana with a side length of 42 cm. What would
you expect the circumference of the other dog’s
neck to be? Round to the nearest centimeter.
Tessa needs a 45°-45°-90° triangle with a
hypotenuse of 42 cm.

Example: Using the 30º-60º-90º Triangle Theorem
An ornamental pin is in the shape of an
equilateral triangle. The length of each
side is 6 centimeters. Josh will attach the
fastener to the back along AB. Will the
fastener fit if it is 4 centimeters long?
Step 1 The equilateral triangle is divided into two
30°-60°-90° triangles.
The height of the triangle is the length of the
longer leg.

Example Continued
Step 2 Find the length x of the shorter leg.
Step 3 Find the length h of the longer leg.
The pin is approximately 5.2 centimeters high.
So the fastener will fit.
Hypotenuse = 2(shorter leg)6 = 2x
3 = x Divide both sides by 2.

Check It Out! Example
What if…? A manufacturer wants to make
a larger clock with a height of 30
centimeters. What is the length of each
side of the frame? Round to the nearest
tenth.
Step 1 The equilateral triangle is divided into two
30º-60º-90º triangles.
The height of the triangle is the length of the
longer leg.

Check It Out! Example Continued
Step 2 Find the length x of the shorter leg.
Each side is approximately 34.6 cm.
Step 3 Find the length y of the longer leg.
Rationalize the denominator.
Hypotenuse = 2(shorter leg)y = 2x
Simplify.

Module 6

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