1. Section 5.1 Bisectors of Triangles

2. We learned earlier that a segment bisector is any line, segment, or plane that intersects a segment at its midpoint. If a bisector is also perpendicular to the segment, it is called a perpendicular bisector.

4. Example 1: a) Find the length of BC. BC = ACPerpendicular Bisector Theorem BC = 8.5Substitution

5. Example 1: b) Find the length of XY.

6. Example 1: c) Find the length of PQ. PQ = RQPerpendicular Bisector Theorem 3x + 1 = 5x – 3 Substitution 1 = 2x – 3 Subtract 3x from each side. 4 = 2xAdd 3 to each side. 2 = xDivide each side by 2. So, PQ = 3(2) + 1 = 7

7. When three or more lines intersect at a common point, the lines are called concurrent lines. The point where concurrent lines intersect is called the point of concurrency. A triangle has three sides, so it also has three perpendicular bisectors. These bisectors are concurrent lines. The point of concurrency of the perpendicular bisectors is called the circumcenter of the triangle.

8. The circumcenter can be on the interior, exterior, or side of a triangle.

9. Example 2: GARDEN A triangular-shaped garden is shown. Can a fountain be placed at the circumcenter and still be inside the garden? By the Circumcenter Theorem, a point equidistant from three points is found by using the perpendicular bisectors of the triangle formed by those points.

10. No, the circumcenter of an obtuse triangle is in the exterior of the triangle. Copy ΔXYZ, and use a ruler and protractor to draw the perpendicular bisectors. The location for the fountain is C, the circumcenter of ΔXYZ, which lies in the exterior of the triangle. C

11. We learned earlier that an angle bisector divides an angle into two congruent angles. The angle bisector can be a line, segment, or ray.

13. Example 3: a)Find the length of DB. DB= DCAngle Bisector Theorem DB= 5Substitution

14. Example 3: b) Find m  WYZ.  WYZ   XYWDefinition of angle bisector m  WYZ= m  XYWDefinition of congruent angles m  WYZ= 28°Substitution

15. Example 3: c) Find the length of QS. QS= SRAngle Bisector Theorem 4x – 1= 3x + 2Substitution x – 1= 2Subtract 3x from each side. x= 3Add 1 to each side. So, QS = 4(3) – 1 or 11.

16. The angle bisectors of a triangle are concurrent, and their point of concurrency is called the incenter of a triangle.

17. Example 4: a) Find ST if S is the incenter of ΔMNP. By the Incenter Theorem, since S is equidistant from the sides of ΔMNP, ST = SU. Find ST by using the Pythagorean Theorem. a 2 + b 2 = c 2 Pythagorean Theorem 8 2 + SU 2 = 10 2 Substitution 64 + SU 2 = 1008 2 = 64, 10 2 = 100

18. Since length cannot be negative, use only the positive square root, 6. Since ST = SU, ST = 6. SU 2 = 36Subtract 64 from each side. SU = ±6Take the square root of each side.

19. Example 4: b) Find m  SPU if S is the incenter of ΔMNP. Since MS bisects  RMT, m  RMT = 2m  RMS. So m  RMT = 2(31) or 62°. Likewise, m  TNU = 2m  SNU, so m  TNU = 2(28) or 56°. m  UPR + m  RMT + m  TNU =180 ° Triangle Angle Sum Theorem m  UPR + 62 ° + 56 ° =180 ° Substitution m  UPR + 118 ° =180 ° Simplify. m  UPR =62 ° Subtract 118 ° from each side.

20. Since PS bisects  UPR, 2m  SPU = m  UPR. This means that m  SPU = m  UPR. __ 1 2 m  SPU = (62) or 31° __ 1 2