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EXAMPLE 3

City Travel. 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.

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EXAMPLE 3

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  1. City Travel In the diagram, 1, 2, and 3 are all congruent and GF = 120 yards, DE = 150 yards, and CD = 300 yards. Find the distance HFbetween Main Street and South Main Street. Corresponding angles are congruent, so FE,GD, and HCare parallel. Use Theorem 6.6. EXAMPLE 3 Use Theorem 6.6 SOLUTION

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

  3. In the diagram, QPRRPS. 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 SOLUTION

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

  5. Find the length of AB. 3. 4 2 ANSWER 19.2 ANSWER 4. for Examples 3 and 4 GUIDED PRACTICE

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