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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 
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   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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