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: )
I tried to prove it without any success. Can someone please give me a hint?
$\endgroup$2 Answers
$\begingroup$The incenter point of a triangle is the intersection of its (interior) angle bisectors and its exitence is gurantee by Ceva's theorem.
Since each point of an angle bisector line is equidistant from the two sides of the angle, the incenter is the center of the incircle.
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.
$\endgroup$1
