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EXAMPLE 1

BD is the perpendicular bisector of AC . Find AD. 3 x + 14. 5 x =. x =. 7. 5 x =. 35. AD =. 5(7) =. EXAMPLE 1. Use the Perpendicular Bisector Theorem. ALGEBRA. AD = CD. Perpendicular Bisector Theorem. Substitute. Solve for x.

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EXAMPLE 1

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  1. BDis the perpendicular bisector of AC. Find AD. 3x + 14 5x= x = 7 5x = 35. AD = 5(7) = EXAMPLE 1 Use the Perpendicular Bisector Theorem ALGEBRA AD=CD Perpendicular Bisector Theorem Substitute. Solve for x.

  2. Because VY = VZ,Vis equidistant from Yand Z. So, by the Converse of the Perpendicular Bisector Theorem, Vis on the perpendicular bisector of YZ, which is WX. In the diagram, WXis the perpendicular bisector of YZ. a. What segment lengths in the diagram are equal? b. IsV onWX ? SOLUTION a. WXbisects YZ , so XY =XZ. Because Wis on the perpendicular bisector of YZ, WY =WZby Theorem 5.2. The diagram shows that VY = VZ = 25. EXAMPLE 2 Use perpendicular bisectors b.

  3. In the diagram, JKis the perpendicular bisector of NL. 1. What segment lengths are equal? Explain your reasoning. SOLUTION NJ =LJsince JKbisects NL. NK = LK by the Perpendicular Bisector Theorem and the diagram shows ML =MN. for Examples 1 and 2 GUIDED PRACTICE

  4. In the diagram, JKis the perpendicular bisector of NL. 2. Find NK. x = 3 x Substitute in 6x – 5 = 3 for Examples 1 and 2 GUIDED PRACTICE SOLUTION NK = LK Perpendicular Bisector Theorem 6x – 5 = 4x + 1 Substitute. Solve for x. We get NK = 13

  5. In the diagram, JKis the perpendicular bisector of NL. Explain why Mis on JK. 3. SOLUTION Since ML = MN, Mis equidistant from Nand L, so by the Converse of the Perpendicular Bisector Theorem Mis on the perpendicular bisector of NLwhich is JK. for Examples 1 and 2 GUIDED PRACTICE

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