If a, b and c are the sides of a right angle triangle where C is the hypotenuse. Prove that radius, r of the circle touches the sides of the triangle given by ** r=a+b-c **

** 2**

Let the circle touches the sides BC, CA, AB of the right triangle ABC at D, E and F respectively, where BC = a, CA = b and AB = c Then AE = AF and BD = BF. Also CE = CD = r.

i.e., b – r = AF, a – r = BF

or AB = c = AF + BF = b – r + a – r

This gives **r=(a+b-c)/2**