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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.

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Presentation on theme: "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."— Presentation transcript:

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. ANSWER 19.2 4. ANSWER 2 4


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