Bell Ringer

Proportions and Similar Triangles

Example 1 Find Segment Lengths

Find the value of x.

48

x12= Substitute 4 for CD, 8 for DB, x

for CE, and 12 for EA.

4 · 12 = 8 · x Cross product property48 = 8x Multiply.488 = 8x

8 Divide each side by 8.

SOLUTION

CDDB = CE

EA Triangle Proportionality Theorem

6 = x Simplify.

Example 2 Find Segment Lengths

Find the value of y.

39

y20 – y= Substitute 3 for PQ, 9 for QR, y

for PT, and (20 – y) for TS.

3(20 – y) = 9 · y Cross product property

60 – 3y = 9y Distributive property

PQQR = PT

TS Triangle Proportionality Theorem

SOLUTION

You know that PS = 20 and PT = y. By the Segment Addition Postulate, TS = 20 – y.

Example 2 Find Segment Lengths

6012 = 12y

12 Divide each side by 12.

60 – 3y + 3y = 9y + 3y Add 3y to each side.

60 = 12y Simplify.

5 = y Simplify.

Example 3 Determine Parallels

Given the diagram, determine whether MN is parallel to GH.

SOLUTION

Find and simplify the ratios of the two sides divided by MN.

LMMG = 56

21 = 83

LNNH = 48

16 = 31

ANSWER Because ≠ 31

83

, MN is not parallel to GH.

Now You Try Find Segment Lengths and Determine Parallels

Find the value of the variable.

1.

2.

ANSWER 8

ANSWER 10

Checkpoint Find Segment Lengths and Determine Parallels

4.

3.

Given the diagram, determine whether QR is parallel to ST. Explain.

ANSWER

Converse of the Triangle Proportionality Theorem.

=6

1248

Yes; || so QR ST by the

≠1723

1521

no;ANSWER

Now You Try

A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.

Example 4 Use the Midsegment Theorem

Find the length of QS.

12QS = PT = (10) = 51

2

ANSWER The length of QS is 5.

SOLUTION

From the marks on the diagram, you know S is the midpoint of RT, and Q is the midpoint of RP. Therefore, QS is a midsegment of PRT. Use the Midsegment Theorem to write the following equation.

Checkpoint Use the Midsegment Theorem

Find the value of the variable.

ANSWER 8

ANSWER 24

5.

6.ANSWER 28

7. Use the Midsegment Theorem to find the perimeter of ABC.

Now You Try

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