Question

The sides a, b, c of a right triangle, where c is the hypotenuse, are circumscribing a circle. Prove that the radius r of the circle is given by r = (a+b-c) /2


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Answer 1

Answer:

Let the circle touches the sides AB, BC & CA of triangle ABC at D,E & F

Since lengths of tangents drawn from an external point are equal

We have

AD=AF, BD=BE and CE=CF

Similarly EB=BD=r

Then we have

=>c=AF+FC

=>c=AD+CE

=>c=(AB-DB)(CB-EB)

=>c=a-r+b-r

=>2r=a+b-c

r=(a+b-c)/2