Question

Prove that : $a(r{r}_1+r_2{r}_3)=b(r{r}_2+r_3{r}_1)=c(r{r}_3+r_1{r}_2)$

where $r$ is inradius and $r_1,r_2,r_3$ are exradius of triangle $ABC$ and $a,b,c$ are the corresponding sides.

Answer 1

We have;

To prove: $a(r{r}_1+r_2{r}_3)=b(r{r}_2+r_3{r}_1)=c(r{r}_3+r_1{r}_2)$

Formula to be used are:

$r=\frac{\Delta }{s};r_1=\frac{\Delta }{s-a};r_2=\frac{\Delta }{s-b};r_3=\frac{\Delta }{s-c}$

$a(r{r}_1+r_2{r}_3)=a[\frac{\Delta }{s}\times \frac{\Delta }{(s-a)}+\frac{\Delta }{(s-b)}\times \frac{\Delta }{(s-c)}]$

$=a\left[{\Delta }^2\left(\frac{(s-b)(s-c)+s(s-a)}{s(s-a)(s-b)(s-c)}\right)\right][\because \Delta =\sqrt{s(s-a)(s-b)(s-c)}]$

$=a[s^2-(a+b+c)s+bc+s^2]$

$[\because 2s=a+b+c]$

$=a[2s^2-2s.s+bc]$

$a(r{r}_1+r_2+r_3)=abc$ (1)

$b(r{r}_2+r_3.r_1)=b[\frac{\Delta }{s}.\frac{\Delta }{(s-a)}+\frac{\Delta }{(s-b)}\times \frac{\Delta }{(s-c)}]$

$=b\left[{\Delta }^2\left(\frac{(s-c)(s-a)+s(s-b)}{{\Delta }^2}\right)\right]$

$=b[2s^2-(a+b+c)s+ac]$

$=b[2s^2-2s^2+ac]$

$b(r{r}_2+r_3{r}_1=abc)$ (2)

$c(r{r}_3+r_1{r}_2)=c[\frac{\Delta }{s}.\frac{\Delta }{s-c}+\frac{\Delta }{s-a}.\frac{\Delta }{s-a}]$

$=c\left[{\Delta }^2\left(\frac{(s-a)(s-b)+s(s-c)}{{\Delta }^2}\right)\right]$

$=c[2s^2-(a+b+c)s+ab]$

$=c[2s^2-2s^2+ab]$

$c(r{r}_3+r_1{r}_2)=abc$ (3)

$\therefore$ equation 1=equation 2 = equation 3

Hence; $a(r{r}_1+r_2{r}_3)=b(r{r}_2+r_1{r}_3)=c(r{r}_3+r_1{r}_2)$