1. Section 5.2: Bisectors of a Triangle

2. Perpendicular bisector of a triangle – A line, ray, or segment that is perpendicular to a side of the triangle at the midpoint of the side.

3. Concurrent lines – when three or more lines, rays, or segments intersect in the same point. Point of concurrency – the point of intersection of the lines, rays, or segments.

4. Circumcenter of the triangle – the point of concurrency of the perpendicular bisectors of a triangle.

5. Theorem 5.5: Concurrency of Perpendicular Bisectors of a Triangle The perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle. Circumcenter is point P PA = PB = PC

6. Angle bisector of a triangle – a bisector of an angle of the triangle. Incenter of the triangle – the point of concurrency of the angle bisectors.

7. Theorem 5.6: Concurrency of Angle Bisectors of a Triangle The angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle. (diagram on pg. 274) PD = PE = PF

8. The Pythagorean Theorem: In any right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. a 2 + b 2 = c 2 a c b

9. Example 1: Three people need to decide on a location to hold a monthly meeting. They will all be coming from different places in the city, and they want to make the meeting location the same distance from each person. A B C Explain why using the circumcenter as the location for the meeting would be the fairest for all. Because the circumcenter is equidistant from the three vertices. Each person would have the exact same distance to travel.

10. Example 2: DG, EG, and FG are the perpendicular bisectors of ΔABC. Find GC. GC = 13.4

11. Example 3: The angle bisectors of ∆MNP meet at point L. What segments are congruent? Find LQ and LR. ML = 17 MQ = 15

12. By Theorem 5.6, the three angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle. So, LR  LQ  LS

13. Use the Pythagorean Theorem to find LQ in ∆LQM M a 2 + b 2 = c 2 Q L (LQ) 2 + (MQ) 2 = (LM) 2 (LQ) 2 + (15) 2 = (17) 2 (LQ) 2 + 225 = 289 (LQ) 2 = 64 LQ = 8 Because LR  LQ, LR = 8

14. HOMEWORK pg. 275 – 276; 3 – 4, 10 – 17