$\begin{aligned} & \text {} \because \angle C=90^{\circ} \Rightarrow C^2=a^2+b^2 \\ & \text { Now, }\left(\frac{r_1-r_3}{r_1}\right)\left(\frac{r_2-r_3}{r_2}\right)=\frac{\frac{\Delta}{s-a}-\frac{\Delta}{s-c}}{\frac{\Delta}{s-a}} \cdot \frac{\frac{\Delta}{s-b}-\frac{\Delta}{s-c}}{\frac{\Delta}{s-b}} \\ & =\frac{s-c-(s-a)}{s-c} \cdot \frac{s-c-(s-b)}{s-c} \\ & =\frac{a-c}{s-c} \frac{b-c}{s-c}=\frac{a b-a c-b c+c^2}{\left(\frac{a+b+c}{2}-c\right)^2} \\ & =4 \frac{\left(a b-a c-b c+a^2+b^2\right)}{(a+b-c)^2} \\ & =4 \cdot \frac{\left(a b-a c-b c+a^2+b^2\right)}{a^2+b^2+c^2+2 a b-2 b c-2 c a} \\ & =4 \cdot \frac{a^2+b^2+a b-a c-b c}{2\left(a^2+b^2\right)+2 a b-2 b c-2 c a}=2 \quad\left\{\therefore c^2=a^2+b^2\right\}\end{aligned}$