Incenter Equidistant from Sides of TriangleDate: 11/18/2001 at 22:35:10 From: April Russell Subject: Geometry Prove that the point of intersection of the angle bisectors of a triangle is equidistant from the sides of the triangle. Date: 11/19/2001 at 05:08:35 From: Doctor Floor Subject: Re: Geometry Hi, April, Thanks for writing. The key to this problem is that each point on the angle bisector is equidistant from the legs of the angle. We see that D is on the angle bisector of angle A, and segments DE and DF are perpendicular to the two legs of angle A. Now we have that angles EAD and FAD are congruent and that of course in triangles EAD and FAD the hypotenuses AD are congruent. That means that right triangles EAD and FAD are congruent, and thus DE = DF. Now we know that all points on the angle bisector of angle A in triangle ABC are equidistant from AC and AB. But the points on the angle bisector of angle B are equidistant from BC and AB, and that means that the point of intersection is equidistant from AC and AB, but also from BC and AB. In short, this point is equidistant from all three sides. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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