If a,b,c are the sides of a right triangle with c as the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by $r = \frac{a + b - c}{2}$

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Solution

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 $r = \frac{(a + b - c)}{2}$

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If a,b,c are the sides of a right triangle with c as the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r=a+b−c2

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

If a,b,c are the sides of a right triangle with c as the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by $r = \frac{a + b - c}{2}$

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 $r = \frac{(a + b - c)}{2}$