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Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse

Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse. Find each measure. TU. TU = UV.  Bisector Thm. 3 x + 9 = 7 x – 17. Substitute the given values. 9 = 4 x – 17. Subtract 3x from both sides. 26 = 4 x. Add 17 to both sides. 6.5 = x.

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Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse

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  1. Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. TU TU = UV  Bisector Thm. 3x + 9 = 7x – 17 Substitute the given values. 9 = 4x – 17 Subtract 3x from both sides. 26 = 4x Add 17 to both sides. 6.5 = x Divide both sides by 4. So TU = 3(6.5) + 9 = 28.5.

  2. Since DG = EG and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem. Check It Out! Example 1b Find the measure. Given that DE = 20.8, DG = 36.4, and EG =36.4, find EF. DE = 2EF Def. of seg. bisector. 20.8 = 2EF Substitute 20.8 for DE. 10.4 = EF Divide both sides by 2.

  3. Since, JM = LM, and , bisects JKL by the Converse of the Angle Bisector Theorem. Example 2C: Applying the Angle Bisector Theorem Find mMKL. mMKL = mJKM Def. of  bisector 3a + 20 = 2a + 26 Substitute the given values. a + 20 = 26 Subtract 2a from both sides. a= 6 Subtract 20 from both sides. So mMKL = [2(6) + 26]° = 38°

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