Question

With usual notations, prove that in a triangle $ABC,$

$\frac{b-c}{r_1}+\frac{c-a}{r_2}+\frac{a-b}{r_3}=0$

Answer 1

Formulae related to radii of Ex-circles:

$r_1=\frac{△}{s-a},r_2=\frac{△}{s-b},r_3=\frac{△}{s-c}$

$r_1=s\tan \frac{A}{2},$ $r_2=s\tan \frac{B}{2},$ $r_3=s\tan \frac{C}{2}$

We know that $r_1=\frac{area}{s-a}$
Similarly $r_2=\frac{area}{s-b}$ and $r_3=\frac{area}{s-c}$
So $\frac{b-c}{r_1}+\frac{c-a}{r_2}+\frac{a-b}{r_3}=\frac{1}{area}\left[\right(s-a\left)\right(b-c)+(s-b\left)\right(c-a)+(s-c\left)\right(a-b\left)\right]$
On solving, right hand side $=\frac{1}{area}\left[s\right(b-c)-a(b-c)+s(c-a)-b(c-a)+s(a-b)-c(a-b\left)\right]$
$=\frac{1}{area}\left[s\right(b-c+c-a+a-b)-ab+ac-bc+ab-ac+bc]=\frac{1}{area}[0-0]=0$

Hence, $\frac{b-c}{r_1}+\frac{c-a}{r_2}+\frac{a-b}{r_3}=0$