Associated Topics || Dr. Math Home || Search Dr. Math

### Incenter Equidistant from Sides of Triangle

```
Date: 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
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/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search