![Example 2C: Applying the Angle Bisector Theorem
Find mMKL.
, bisects JKL
Since, JM = LM, and
by the Converse of the Angle
Bisector Theorem.
mMKL = mJKM
3a + 20 = 2a + 26
a + 20 = 26
a = 6
Def. of bisector
Substitution.
Subtraction POE
Subtraction POE
So mMKL = [2(6) + 26]° = 38°](https://image.slidesharecdn.com/bisectorandcentroidofatriangle-230103141703-60e09fb4/85/Bisector_and_Centroid_of_a_Triangle-ppt-8-320.jpg)
Bisector_and_Centroid_of_a_Triangle.ppt
- 1. Prove and apply theorems about perpendicular bisectors. Prove and apply theorems about angle bisectors. 5.1 Objectives
- 3. Example 1A: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. MN MN = LN MN = 2.6 Bisector Thm. Substitution
- 4. Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse TU Find each measure. So TU = 3(6.5) + 9 = 28.5. TU = UV Bisector Thm. 3x + 9 = 7x – 17 9 = 4x – 17 26 = 4x 6.5 = x Subtraction POE Addition POE. Division POE. Substitution
- 5. Check It Out! Example 1b Given that DE = 20.8, DG = 36.4, and EG =36.4, which Theorem would you use to find EF? Find the measure. Since DG = EG and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem.
- 6. Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.
- 7. Example 2A: Applying the Angle Bisector Theorem Find the measure. BC BC = DC BC = 7.2 Bisector Thm. Substitution Find the measure. mEFH, given that mEFG = 50°. Since EH = GH, and , bisects EFG by the Converse of the Angle Bisector Theorem.
- 8. Example 2C: Applying the Angle Bisector Theorem Find mMKL. , bisects JKL Since, JM = LM, and by the Converse of the Angle Bisector Theorem. mMKL = mJKM 3a + 20 = 2a + 26 a + 20 = 26 a = 6 Def. of bisector Substitution. Subtraction POE Subtraction POE So mMKL = [2(6) + 26]° = 38°
- 9. Check It Out! Example 2a Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ. mWYZ = mWYX mWYZ + mWYX = mXYZ mWYZ + mWYZ = mXYZ 2(63°) = mXYZ 126° = mXYZ 2mWYZ = mXYZ
- 11. Prove and apply properties of perpendicular bisectors of a triangle. Prove and apply properties of angle bisectors of a triangle. 5.2 Objectives
- 12. The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex. Helpful Hint
- 13. A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. Every triangle has three medians, and the medians are concurrent.
- 14. The point of concurrency of the medians of a triangle is the centroid of the triangle . The centroid is always inside the triangle. The centroid is also called the center of gravity because it is the point where a triangular region will balance. The length of the segment from the vertex to the centroid is twice the length of the segment from the centroid to the midpoint
- 15. Example 1B: Using the Centroid to Find Segment Lengths In ∆LMN, RS = 5 Find SL and RL. . SL = 10 and RL = 15
- 16. Check It Out! Example 1a In ∆JKL, ZK = 14, Find ZW and WK ZW = 7 and WK = 21
- 17. Check It Out! Example 1b In ∆JKL, JY = 36, Find JZ and ZY. JZ = 24 and ZY = 12
- 18. Lesson Drill Use the figure for Items 1–3. In ∆ABC, AE = 12, DG = 7, and BG = 9. Find each length. 1. AG 2. GC 3. GF 8 14 13.5