Problem Statement:The sides of a triangle ABC are AB = 15 cm, BC = 18 cm, and CA = 24 cm. Find the distance from the point of intersection of the angle bisectors to side AB.
The distance from the point of intersection of the angle bisectors to side AB is 4.73 cm. Note that by definition the point of intersection of all angle bisectors of a triangle is called incenter. It is also defined as the center of the inscribed circle of the triangle. Thus, the distance from incenter of the triangle to any of its sides is equal to the radius of the inscribed circle.Solution: