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 the segments of tangents from the vertices to the incircle be $x,y,z$ (what is popularly known as the Ravi-substitution.
We have $(x+y), (y+z), (z+x)$ are all integers and so is $2(x+y+z)$ so $x+y+z = \frac{n}{2}, n \in \mathbb{N}$. Now its easy to see that $x,y,z$ are in the form $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$. Since $x+y, y+z, z+x \in \mathbb{N}, a,b,c$ have the same parity 
We are given that $r=\frac{xyz}{x+y+z} =1 $ i.e. $abc = 4(a+b+c)$ from which its clear that $a,b,c$ are all even $\Rightarrow x,y,z \in \mathbb{N}$.
So we are looking for solutions in $\mathbb{N}$ for $xyz = x+y+z$.
WLOG $x \le y \le z$. Hence $xyz = x+y+z \le 3z \Rightarrow xy \le 3$
Now taking the cases $(x,y) = (1,1), (1,2), (1,3)$ we see that $x=1, y=2, z=3$ is the only solution.
This gives the sides as $3,4,5$ which is a Pythagorean Triplet and hence sides of a right angled triangle.
