Date: Dec 28, 1997 12:01 AM
Author: Eileen M. Klimick Schoaff
Subject: Re: to prove: rays bisecting 3 angles of a triangle meet at a single<br> point
The (1) proof you described in the point at which the medians intersect -- the

center of gravity of the triangle. If you connect this point to the three

vertices, you will have 3 triangles of equal area. If you cut out the

triangle, you can balance the triangle on a pencil point at this point.

The (2) proof is the circumcenter. By bisecting the sides, you are locating a

point equidistant from the 3 vertices -- which would be the center of a circle

going through the 3 vertices.

The proof that you claim you cannot do is to prove that the three angle

bisectors meet at a point. This point is, in fact, called the incenter because

it is the center of the inscribed angle. Consider triangle ABC. If ray AD

bisects angle A, then every point on that ray is equidistant from the sides of

angle A, segments AB and AC. If ray BE bisects angle B, then every point on

that ray is equidistant from the sides of angle B, segments BA and BC. Let O

be the point where the rays BE and AD intersect. Then point O is equidistant

from the 3 sides of the triangle -- AB, BC, AC. Since it is equidistant from

BC and AC, it must lie on the ray that bisects angle C.

If you not construct a perpendicular from O to any side of the triangle, OG,

then you can construct a circle with center O and radius OG that is inscribed

in triangle ABC.

There is a fourth point that you do not mention -- the orthocenter -- that is

the point of intersection of the 3 altitudes of the triangle.

Eileen Schoaff

Buffalo State College