If $a, b, c$ are the sides of a right triangle where $c$ is the hypotenuse,prove that the radius $r$ of the incircle of the triangle is given by $r = \frac{a + b - c}{2}$.

(N/A) Let the incircle of the right triangle $ABC$ touch the sides $BC, CA, AB$ at $D, E, F$ respectively,where $BC = a, CA = b$ and $AB = c$. Let $O$ be the center of the circle and $r$ be its radius.
Since the tangents drawn from an external point to a circle are equal in length,we have $AE = AF$,$BD = BF$,and $CD = CE = r$.
In the quadrilateral $CDOE$,since $\angle C = 90^\circ$ and the radii are perpendicular to the tangents,$\angle ODC = \angle OEC = 90^\circ$. Thus,$CDOE$ is a square with side $r$.
Therefore,$CD = CE = r$.
From the figure,$AF = AE = b - r$ and $BF = BD = a - r$.
Since $AB = c$,we have $AF + BF = c$.
Substituting the values,$(b - r) + (a - r) = c$.
$a + b - 2r = c$.
$2r = a + b - c$.
$r = \frac{a + b - c}{2}$.