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 is equal to $\frac{a+b-c}{2}$.

Answer 1

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
Since length of tangents drawn from an external point are equal.
$\Rightarrow AE=AF$ and $BD=BF$

$OE\perp AC,OD\perp CD$ [$\because$ radius $\perp$ tangents ]

OECD is a square.
So, $CE=CD=r$

and $b-r=AF,a-r=BF$

Therefore, $AB=AF+BF$

$AB=c=AF+BF=(b-r)+(a-r)$

$\Rightarrow c=b-r+a-r$

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

Hence proved.