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Re: to prove: rays bisecting 3 angles of a triangle meet at a single point
Posted:
Dec 28, 1997 12:01 AM


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



