EXAMPLE 1 Use the Perpendicular Bisector Theorem AD = CD Perpendicular Bisector Theorem 3x + 14 5x =5x = Substitute. 7 x = Solve for x. BD is the perpendicular.

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Presentation transcript:

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

EXAMPLE 2 Use perpendicular bisectors a. What segment lengths in the diagram are equal? b. Is V on WX ? SOLUTION a. WX bisects YZ, so XY = XZ. Because W is on the perpendicular bisector of YZ, WY = WZ by Theorem 5.2. The diagram shows that VY = VZ = 25. In the diagram, WX is the perpendicular bisector of YZ. b. Because VY = VZ, V is equidistant from Y and Z. So, by the Converse of the Perpendicular Bisector Theorem, V is on the perpendicular bisector of YZ, which is WX.

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

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

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