```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 -- thecenter of gravity of the triangle.  If you connect this point to the threevertices, you will have 3 triangles of equal area.  If you cut out thetriangle, 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 apoint equidistant from the 3 vertices -- which would be the center of a circlegoing through the 3 vertices.The proof that you claim you cannot do is to prove that the three anglebisectors meet at a point.  This point is, in fact, called the incenter becauseit is the center of the inscribed angle.  Consider triangle ABC.  If ray ADbisects angle A, then every point on that ray is equidistant from the sides ofangle A, segments AB and AC.  If ray BE bisects angle B, then every point onthat ray is equidistant from the sides of angle B, segments BA and BC.  Let Obe 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 fromBC 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 inscribedin triangle ABC.There is a fourth point that you do not mention -- the orthocenter -- that isthe point of intersection of the 3 altitudes of the triangle.Eileen SchoaffBuffalo State College
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