Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.

Let $a$, $b$ and $c$ be sides-lengths of the triangle.
Thus, $$1=\frac{2S}{a+b+c}$$ or
$$4(a+b+c)^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ or
$$4(a+b+c)=(a+b-c)(a+c-b)(b+c-a).$$
Now, it's obvious that $a+b+c$ is an even number.
Thus, $a+b-c$, $a+c-b$ and $b+c-a$ are even positive numbers.
Let $a+b-c=2w$, $a+c-b=2v$ and $b+c-a=2u$.
Thus, we need to solve
$$u+v+w=uvw$$ or
$$\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=1.$$
Let $uv=z$, $uw=y$ and $vw=x$.
Hence, $x$, $y$ and $z$ are natural numbers such that
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$$
This equation we can prove by the following way.
Let $x\leq y\leq z$.
Hence, $$1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq\frac{3}{x},$$
which gives $x\leq3$.
$x=1$ is impossible and $x=2$ and $x=3$ give the following solutions:
$(2,3,6)$, $(2,4.4)$ and $(3,3,3)$.
Last two are impossible and $(x,y,z)=(2,3,6)$ gives $(u,v,w)=(3,2,1)$,
which gives $(a,b,c)=(3,4,5)$ and since $4^2+3^2=5^2,$ we are done!
