Download Geometry Section 5.2 Notes

CHAPTER
5
5.2 Similar Polygons
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Slide 7-1
Similar Polygons
Similar polygons (figures) have the same shape but
not necessarily the same size. We will abbreviate “is
similar to” with the symbol ∼.
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Slide 7-2
Similar Polygons
We write a similarity statement with corresponding
vertices in order, the same way we write a
congruence statement. When three or more ratios
are equal, we can write an extended proportion.
AB BC CD AD
The proportion GH  HI  IJ  GJ
is an
extended proportion.
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Slide 7-3
Understanding Similarity and
Example
Using Extended Proportions
MNP
SRT
a. What are the pairs of congruent angles?
Use the order of the vertices in the similarity
statement ΔMNP ∼ ΔSRT to write pairs of
congruent angles.
∠M ≅ ∠S, ∠N ≅ ∠R, and ∠P ≅ ∠T
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Slide 7-4
Understanding Similarity and
Example
Using Extended Proportions
MNP
SRT
b. What is the extended proportion for the ratios of
corresponding sides?
Since ΔMNP ∼ ΔSRT, we know that MN corresponds to
MN
is a ratio of corresponding sides the same is true
SR, so
SR
for
MN NP MP


SR RT ST
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Slide 7-5
Angle-Angle Similarity (AA ∼ ) Postulate
If two angles of one triangle are congruent to two
angles of another triangle, then the triangles are
similar.
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Slide 7-6
Example
Using the AA ∼ Postulate
Determine whether ΔRSW and ΔVSB are similar.
Explain.
Solution
By studying the diagram,
we see that we can use the AA ∼ Postulate.
Show that two pairs of angles are congruent.
∠R ≅ ∠V because both angles measure 45°.
∠1 ≅ ∠2 because vertical angles are congruent.
So, ΔRSW ∼ ΔVSB by the AA ∼ Postulate.
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Slide 7-7
Example
Using the AA ∼ Postulate
Determine whether ΔJKL and ΔPQR are similar.
Explain.
Solution
Look at the diagram, we see that we can use the
AA ∼ Postulate. Show two pairs of angles congruent.
∠L ≅ ∠R because both angles measure 70°.
m∠K = 180° − 30° − 70° = 80°
m∠P = 180° − 85° − 70° = 25°
Only one pair of angles is congruent. So, ΔJKL and
ΔPQR are not similar.
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Slide 7-8
Theorem 5.11 Triangle Prop. Theorem
Theorem If a line is parallel to one side of a
triangle and intersects the other two sides, then it
divides those sides proportionally.
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Slide 7-9
Example Triangle Prop. Theorem
What is the value of x?
Solution
KL PN a side of triangle PMN.
Set up a proportion using the
Side-Splitter Theorem.
PK
NL

KM LM
9 x  9  12 x
x 1 x
9  3x

12
9
3 x
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Slide 7-10
Transitive Law for Similar Triangles
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Slide 7-11