Circles question on proof

Question

It is given that a, b, and c are the sides of a triangle and c is the hypotenuse. There is an incircle inside the triangle with radius = r.

We need to prove that $r=\dfrac{a+b-c}{2}$

Image:

My attempt

Area of circle = $\dfrac{ab}{2} = \dfrac{ar+br+cr}{2}$ (area of individual triangles)

$r=\dfrac{ba}{a+b+c}$

but we have to prove that $r=\dfrac{a+b-c}{2}$ . How can that be shown? Thank you very much

Answer 1

Quoting math.wichita.edu,

Let r be the inradius. Since the tangents to a circle from a point outside the circle are equal, we have the sides of triangle ABC configured as in the above figure. Thus,

$c = a + b - 2r$

$= r = (a + b - c)/2$

$= c = (a - r) + (b - r)$

Answer 2

Continuing from where you had left, $$r=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{(a+b)^2-c^2}$$ $$\rightarrow r=\frac{ab(a+b-c)}{(a^2+b^2-c^2)+2ab}=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}$$