Question

With usual notations, prove that in a triangle ABC
$\frac{r_1}{(s-b)(s-c)}+\frac{r_2}{(s-c)(s-a)}+\frac{r_3}{(s-a)(s-b)}=\frac{3}{r}$

Answer 1

We know that $r_1=\frac{area}{s-a}$ ,$r_2=\frac{area}{s-b}$,$r_3=\frac{area}{s-c}$

on putting these values in left hand side

$\frac{area}{(s-a)(s-b)(s-c)}\times 3=3\times \frac{area\times s}{are{a}^2}=3\times \frac{s}{area}=3\times \frac{1}{r}=\frac{3}{r}$

$\text{Here we used}:s(s-a)(s-b)(s-c)=are{a}^2$