Type of triangle-2

Let $a$ , $b$ and $c$ be the sides of $\triangle ABC$ . Now, we can express the three ex-radii with the following formulas (with $r_1$ opposite to $A$ and so on), where $[ABC]$ is the area of $\triangle ABC$ :

$r_1=\dfrac{2[ABC]}{-a+b+c} , r_2=\dfrac{2[ABC]}{a-b+c} , r_3=\dfrac{2[ABC]}{a+b-c}$

Now, substituting in the given expression:

$\left(1-\dfrac{a-b+c}{-a+b+c}\right)\left(1-\dfrac{a+b-c}{-a+b+c}\right)=2$

Now try to simplify until we obtain something familiar:

$\left(\dfrac{-a+b+c-a+b-c}{-a+b+c}\right)\left(\dfrac{-a+b+c-a-b+c}{-a+b+c}\right)=2 \\ \dfrac{2(b-a)}{-a+b+c} \cdot \dfrac{2(c-a)}{-a+b+c}=2 \\ 2(b-a)(c-a)=(-a+b+c)^2 \\ 2bc-2ab-2ac+2a^2=a^2+b^2+c^2-2ab-2ac+2bc \\ a^2=b^2+c^2$

By the converse of the Pythagoeran Theorem, $\triangle ABC$ is a right triangle at $A$ .