Chapter 5Section 5.2

Bisectors of Triangles

Theorem 5.5 Concurrency of the Perpendicular Bisectors of a triangle

Theorem

Theorem 5.5 Concurrency of the Perpendicular Bisectors of a triangle

The perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.

A

C B

DThe circumcenter is equidistant from the

vertices of the triangle

Theorem 5.6 Concurrency of the Angle Bisectors of a triangle

Theorem

Theorem 5.6 Concurrency of the Angle Bisectors of a triangle

The angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.

The incenter is equidistant from the sides of the triangle

D is the circumcenter so it is equidistant from all the vertices of the triangle

DA = 2

Since D is equidistant from A and B;

AB = 4

A B, BED DFA,

,DABD A.A.S.

V is the incenter so it is equidistant from all the sides of the triangle

VS = 3

mVZX = 20

S.A.S.

SZ Bisects WZX

XT Bisects WXZ

WZX WXZ

SVX TVZ

,VTSV ZVXV

Sometimes

Always

Always

Never

This distance can be found by using the Pythagorean

Theorem

Since the incenter is equidistant from each side

of the triangle, this distance is the same

82 + x2 = 102

64 + x2 = 100

x2 = 36

x = 6 ID = 6

DB = AD

DB = 15

These two triangles are congruent by SAS

HW #55Pg 275-278 10-19, 20-22, 24-28, 32-40