Question

in the given fig. n, l, and m are the length of the side LM, MN, and NL, of right angled triangle LMN...If r is the radius of the incircle, then Prove that 2r=(n+l-m)

Answer 1

Let the circle touches the sides MN, LM, LN of the right triangle ABC at G, E and F respectively, where MN= a, LM = b and LN = 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
2r = a + b - c.

Answer 2

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