$\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$





$$
\begin{aligned}
& \text { Now, } r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c} \\
& \therefore\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-b}}\right)\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-c}}\right) \\
& =\left(1-\frac{s-b}{s-a}\right)\left(1-\frac{s-c}{s-a}\right)=\frac{(b-a)(c-a)}{(s-a)^2}=2 \\
& \therefore b c+a^2-a b-a c=\frac{1}{2}(2 s-2 a)^2 \\
& \Rightarrow 2 b c+2 a^2-2 a b-2 a c=(b+c-a)^2 \\
& =b^2+c^2+a^2-2 a b-2 a c+2 b c \\
& \Rightarrow \quad a^2=b^2+c^2
\end{aligned}
$$
$\therefore A B C$ is right angled triangle.