1. 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.

2. HG + GF GF CD + DE DE = EXAMPLE 3 HG GF CD DE = Parallel lines divide transversals proportionally. Property of proportions (Property 4 ) Substitute. HF = 360 HF 120 300+150 150 = 450 150 HF 120 = Simplify. Multiply each side by 120 and simplify. The distance between Main Street and South Main Street is 360 yards. Use Theorem 6.6

3. 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.

4. EXAMPLE 4 Use Theorem 6.7 7x = 195 – 13x Cross Products Property x = 9.75 Solve for x. RQ RS PQ PS = 15 – x x = 7 3 Angle bisector divides opposite side proportionally. Substitute.

5. 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.

6. GUIDED PRACTICE for Examples 3 and 4 Parallel lines divide transversals proportionally. AB 16 18 15 = AB = 19.2 Simplify. So, the length of AB is 19.2 ANSWER

7. 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.

8. GUIDED PRACTICE for Examples 3 and 4 2 x = 4 Solve for x. ANSWER The length of x = 2 4 Substitute. = 4 4 4 2 x 4x = 4 2 Cross Products Property