Prove by geometry or algebra

Question

$a, b$ and $c$ are the sides of right triangle where $c$ is the hypotenuse. Show that the inradius, $r$, of the right triangle satisfies $a+b+c=2r$. Circle is inscribed into the triangle.

Answer 1

If you have right triangle $ABC$ and inscribed cirle touches sides $AB, BC, CA$ at points $M,N,K$ respectively, then we know that $CK=CN$, $BN=BM$, $AM=AK \ \implies CK=CN = \frac{AB+BC-CA}{2} \implies \frac{a+b-c}{2}=CK=CN=r \implies 2r = a +b - c$

Answer 2

Above I posted geometrical solution. Also we can prove it using algebra:

Indeed, we know that

$Area = \frac{a+b+c}{2} \times r \implies 2r = \frac{4\times Area}{a+b+c} = \frac{2ab}{a+b+c} = \frac{2ab(a+b-c)}{(a+b+c)(a+b-c)}=$

$=\frac{2ab(a+b-c)}{a^2+b^2+2ab-c^2}=\frac{2ab(a+b-c)}{2ab}=a+b-c$.