Question

In triangle ABC, prove that rr1=(s-b)(s-c)​

Answer 1

Given:

We have to prove that  $rr_1=(s-b)(s-c)$

Solution:

We have the relations,

$r=\frac{\triangle}{s}$

$r_1=\frac{\triangle}{s-a}$

and ${\triangle}^{2}=s(s-a)(s-b)(s-c)$

Here, L.H.S = $rr_1$

                   = $(\frac{\triangle}{s})(\frac{\triangle}{s-a} )$

                   = $\frac{\triangle^{2} }{s(s-a)}$

                   = $\frac{s(s-a)(s-b)(s-c)}{s(s-a)}$

       L.H.S   = $(s-b)(s-c)$  =  R.H.S

∴ $rr_1=(s-b)(s-c)$

Hence verified.