Question

a, b and c are the sides of a right triangle, where c is the hypotenuse. A circle, of radius r, touches the sides of the triangle. Prove that
$r = \frac{a + b - c}{2}$
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Answer 1

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

Answer 2

Step-by-step explanation:

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