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Do Now

Do Now. 1. Find the value of n. 2. ∆ XYZ is the midsegment triangle of ∆ WUV. What is the perimeter of ∆ XYZ?. 16. 8. 19. 5. 12. 5.2 Perpendicular and Angle Bisector. Target: SWBAT use properties and apply theorems about perpendicular and angle bisectors.

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Do Now

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  1. Do Now 1. Find the value of n. 2.∆XYZ is the midsegment triangle of∆WUV. What is the perimeter of∆XYZ? 16 8 19 5 12

  2. 5.2 Perpendicular and Angle Bisector Target: SWBAT use properties and apply theorems about perpendicular and angle bisectors.

  3. When a point is the same distance from two or more objects, the point is said to be equidistantfrom the objects. Triangle congruence theorems can be used to prove theorems about equidistant points.

  4. Example 1A: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. MN MN = LN  Bisector Thm. MN = 2.6 Substitute 2.6 for LN.

  5. Since AB = AC and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem. Example 1B: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. BC BC = 2CD Def. of seg. bisector. BC = 2(12) = 24 Substitute 12 for CD.

  6. 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.

  7. Example 2A: Applying the Angle Bisector Theorem Find the measure. BC BC = DC  Bisector Thm. BC = 7.2 Substitute 7.2 for DC.

  8. Since EH = GH, and , bisects EFGby the Converse of the Angle Bisector Theorem. Example 2B: Applying the Angle Bisector Theorem Find the measure. mEFH, given that mEFG = 50°. Def. of  bisector Substitute 50° for mEFG.

  9. 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°

  10. Use this diagram for Items 3–4. 3. Given that FH is the perpendicular bisector of EG, EF = 4y – 3, and FG = 6y – 37, find FG. 4. Given that EF = 10.6, EH = 4.3, and FG = 10.6, find EG. Lesson Quiz: Part I Use this diagram for Items 1–2. 1. Given that mABD= 16°, find mABC. 2. Given that mABD= (2x + 12)° and mCBD= (6x – 18)°, find mABC. 32° 54° 65 8.6

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