In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Let BQ and CP be the bisectors of ∠ABC and ∠ACB respectively, intersecting in the interior of ΔABC at R. Let BQ intersect side AC in Q and CP intersect side AB in P. ∴By angle bisector theorem, Since, R lies on BQ, point R is equidistant from AB and BC. Similarly, R lies on CP and is equidistant from AC and BC. So, O is equidistant from BC and AC. Therefore, point O is equidistant from all three sides AB, BC and CA of ΔABC.