Question

a, b and C are the sides of a right triangle, where c is the hypotenuse. A circle of radius r is, touches the side of triangle.
Prove that
$r \: = \: (a + b + c ) \div 2$

Answer 1

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

HIII......

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