Perpendicular from incenter of a triangle to any side is equal to the radius of the incircle Given a triangle $ABC$ with incenter $I$, it is said that the perpendicular line segment from $I$ to any of the sides $AB$, $AC$, or $BC$ is equal to the radius of the incircle. (See the second picture on this page: http://mathworld.wolfram.com/RightTriangle.html )
I tried to prove it without any success. Can someone please give me a hint?

Well the definition of an incenter is the center of the largest circle that fits into the triangle. So the circle is externally tangent to each side of the triangle.  A well-known circle theorem is that the radius at the point where a tangent touches the circle is perpendicular to the tangent.
