Experts solve this

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  (lengths of tangents from an external points are equal)
 

Also CE = CD = r.  (Since OECD is a square as adjacent sides are equal and perpendicular to each other)

So,
 

 b – r = AF,
 

 a – r = BF

Therefore,
 

AB = c = AF + BF = b – r + a – r
 

This gives

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

Hence proved.