5-3 Use Angle Bisectors of Triangles Warm Up Lesson Presentation

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Learning Targets Prove and apply theorems about perpendicular bisectors. Prove and apply theorems about angle bisectors.

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Presentation transcript:

5-3 Use Angle Bisectors of Triangles Warm Up Lesson Presentation Lesson Quiz Holt Geometry

Warm Up Construct each of the following. 1. A perpendicular bisector. 2. An angle bisector. 3. Find the midpoint and slope of the segment (2, 8) and (–4, 6).

Objectives Prove and apply theorems about angle bisectors.

Based on these theorems, an angle bisector can be defined as the locus of all points in the interior of the angle that are equidistant from the sides of the angle.

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

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

Example 2C: Applying the Angle Bisector Theorem Find mMKL. , bisects JKL Since, JM = LM, and by the Converse of the Angle Bisector Theorem. 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°

Check It Out! Example 2a Given that YW bisects XYZ and WZ = 3.05, find WX. WX = WZ  Bisector Thm. WX = 3.05 Substitute 3.05 for WZ. So WX = 3.05

Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ. Check It Out! Example 2b Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ. mWYZ + mWYX = mXYZ  Bisector Thm. mWYZ = mWYX Substitute m WYZ for mWYX . mWYZ + mWYZ = mXYZ 2mWYZ = mXYZ Simplify. 2(63°) = mXYZ Substitute 63° for mWYZ . 126° = mXYZ Simplfiy .

Application John wants to hang a spotlight along the back of a display case. Wires AD and CD are the same length, and A and C are equidistant from B. How do the wires keep the spotlight centered? It is given that . So D is on the perpendicular bisector of by the Converse of the Angle Bisector Theorem. Since B is the midpoint of , is the perpendicular bisector of . Therefore the spotlight remains centered under the mounting.