EXAMPLE 3 Use Theorem 6.6

In the diagram, 1, 2, and 3 are all congruent and GF = 120 yards, DE = 150 yards, and CD = 300 yards. Find the distance HF between Main Street and South Main Street.

City Travel

SOLUTION

Corresponding angles are congruent, so FE, GD , and HC are parallel. Use Theorem 6.6.

HG + GF GF

CD + DE DE=

EXAMPLE 3

HGGF

CDDE= Parallel lines divide transversals

proportionally.

Property of proportions (Property 4)

Substitute.

HF = 360

HF120

300+150 150=

450150

HF120 = Simplify.

Multiply each side by 120 and simplify.

The distance between Main Street and South Main Street is 360 yards.

Use Theorem 6.6

SOLUTION

EXAMPLE 4 Use Theorem 6.7

In the diagram, QPR RPS. Use the given side lengths to find the length of RS .

~

Because PR is an angle bisector of QPS, you can apply Theorem 6.7. Let RS = x. Then RQ = 15 – x.

EXAMPLE 4 Use Theorem 6.7

7x = 195 – 13x Cross Products Property

x = 9.75 Solve for x.

RQRS

PQPS=

15 – x x

= 73

Angle bisector divides opposite side proportionally.

Substitute.

GUIDED PRACTICE for Examples 3 and 4

Find the length of AB.

3.

SOLUTION

Corresponding angles are congruent, so the three vertical lines are parallel. Use Theorem 6.6.

GUIDED PRACTICE for Examples 3 and 4

Parallel lines divide transversals proportionally.

AB16

18 15=

AB = 19.2 Simplify.

So, the length of AB is 19.2ANSWER

GUIDED PRACTICE for Examples 3 and 4

4.

SOLUTION

Because DA is an angle bisector of CAB, you can apply Theorem 6.7. Let AB = x.

DC DB

AC AB = Angle bisector divides opposite

side proportionally.

GUIDED PRACTICE for Examples 3 and 4

2x = 4 Solve for x.

ANSWER The length of x = 24

Substitute.=44

4 2x

4x =4 4 2 Cross Products Property