If a, b and c are the sides of a right - angled triangle where c is the hypotenuse then prove that the radius r or the circle which touches the sides of the triangle is given by r = a + b - c/ 2.

The figure is given below:

Let the circle touches the sides BC, CA, AB of the right triangle ABC (right angled at C) at D, E and F respectively,

where BC = a, CA = b , AB = c respectively

Since lengths of tangents drawn from an external point are equal

Therefore, AE = AF,

and BD = BF

Also CE = CD = r and b - r = AF,

a - r = BF

Therefore AB = AF + BF

c = b - r + a - r

AB = c = AF + BF

= b - r + a - r

hence, r =