Question

In a triangle ABC, if $\frac{a-b}{b-c}$=$\frac{s-a}{s-c},$,then $r_1,r_2,r_3$ are in:

• A. $A.P$
• B. $G.P$
• C. $H.P$
• D. None of These

Answer 1

The correct option is A $A.P$

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

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

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

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

$\Rightarrow \frac{\frac{\Delta }{r_1}}{\frac{\Delta }{r_2}-\frac{\Delta }{r_1}}=\frac{\frac{\Delta }{r_3}}{\frac{\Delta }{r_3}-\frac{\Delta }{r_2}}$

$\Rightarrow \frac{\frac{1}{r_1}}{\frac{1}{r_2}-\frac{1}{r_1}}=\frac{\frac{1}{r_3}}{\frac{1}{r_3}-\frac{1}{r_2}}$

$\Rightarrow \frac{\frac{1}{r_1}}{\frac{r_1-r_2}{r_1{r}_2}}=\frac{\frac{1}{r_3}}{\frac{r_2-r_3}{r_2{r}_3}}$

$\Rightarrow \frac{1}{r_1-r_2}=\frac{1}{r_2-r_3}$

$\Rightarrow r_1-r_2=r_2-r_3$

$\Rightarrow 2r_2=r_1+r_3$

$\therefore r_1,r_2,r_3$ are in A.P