![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/perpendicularandanglebisectoractivitylesson-230503073750-6e503452/85/Perpendicular_and_Angle_Bisector_Activity_Lesson-ppt-10-320.jpg)
Perpendicular_and_Angle_Bisector_Activity_Lesson.ppt
- 1. Prove and apply theorems about perpendicular bisectors. Prove and apply theorems about angle bisectors.
- 2. A locus is a set of points that satisfies a given condition. The perpendicular bisector of a segment can be defined as the locus of points in a plane that are equidistant from the endpoints of the segment. When a point is the same distance from two or more objects, the point is said to be equidistant from 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 MN = 2.6 Bisector Thm. Substitution
- 5. 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
- 6. 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.
- 8. Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.
- 9. 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.
- 10. 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°
- 11. 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
- 12. Prove and apply properties of perpendicular bisectors of a triangle. Prove and apply properties of angle bisectors of a triangle.
- 13. The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex. Helpful Hint
- 14. When three or more lines intersect at one point, the lines are said to be concurrent. The point of concurrency is the point where they intersect. In the construction, you saw that the three perpendicular bisectors of a triangle are concurrent. This point of concurrency is the circumcenter of the triangle.
- 15. The circumcenter can be inside the triangle, outside the triangle, or on the triangle.
- 16. The circumcenter of ΔABC is the center of its circumscribed circle. A circle that contains all the vertices of a polygon is circumscribed about the polygon.
- 17. Example 1: Using Properties of Perpendicular Bisectors G is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of ∆ABC. DG, EG, and FG are the perpendicular bisectors of ∆ABC. Find GC. GC = CB GC = 13.4 Circumcenter Thm. Substitute 13.4 for GB.
- 18. Example Find GM. GM = MJ GM = 14.5 MZ is a perpendicular bisector of ∆GHJ. Find GK. GK = KH GK = 18.6 KZ is a perpendicular bisector of ∆GHJ. Find JZ. JZ = GZ JZ = 19.9 Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of ∆GHJ.
- 19. Example 2: Finding the Circumcenter of a Triangle Find the circumcenter of ∆HJK with vertices H(0, 0), J(10, 0), and K(0, 6). Step 1 Graph. Step 2 Find equations for two perpendicular bisectors. The perpendicular bisector of HJ is x = 5, and the perpendicular bisector of HK is y = 3. Step 3 Find the intersection of the two equations. (5, 3) is the circumcenter of ∆HJK.
- 20. Check It Out! Example 2 Find the circumcenter of ∆GOH with vertices G(0, –9), O(0, 0), and H(8, 0) . Step 1 Graph. Step 2 Find equations for two perpendicular bisectors. The perpendicular bisector of GO is y = –4.5, and the perpendicular bisector of OH is x = 4. Step 3 Find the intersection of the two equations. The lines intersect at (4, –4.5), the circumcenter of ∆GOH.
- 21. A triangle has three angles, so it has three angle bisectors. The angle bisectors of a triangle are also concurrent. This point of concurrency is the incenter of the triangle .
- 22. Unlike the circumcenter, the incenter is always inside the triangle.
- 23. The incenter is the center of the triangle’s inscribed circle. A circle inscribed in a polygon intersects each line that contains a side of the polygon at exactly one point.
- 24. Example 3A: Using Properties of Angle Bisectors MP and LP are angle bisectors of ∆LMN. Find the distance from P to MN. MP and LP are angle bisectors of ∆LMN. Find mPMN.