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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
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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
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