N, l and m are the lengths of lm, mn, nl respectively pf a right angled triangle lmn. if r is radius of the incircle ,then prove that 2r =(n+l-m)
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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.
Read more on Brainly.in - https://brainly.in/question/2421566#readmore
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.
Read more on Brainly.in - https://brainly.in/question/2421566#readmore
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