Question

In a $△ABC$, if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1,r_2$ and $r_3$ are

• A. in arithmetic progression
• B. in geometric progression
• C. in harmonic progression
• D. equal

Answer 1

Answer: (A)

$\text{ In }△ABC$
$r_1=\frac{\Delta }{s-a},r_2=\frac{\Delta }{s-b}\text{ and }{r}_3=\frac{\Delta }{s-c}$
$\because (a-b)(s-c)=(b-c)(s-a)$
$\Rightarrow \frac{s-c}{b-c}=\frac{s-a}{a-b}$
$\Rightarrow \frac{s-c}{(s-c)-(s-b)}=\frac{s-a}{(s-b)-(s-a)}$
$\Rightarrow \frac{\frac{\Delta }{r_3}}{\frac{\Delta }{r_3}-\frac{\Delta }{r_2}}=\frac{\frac{\Delta }{r_1}}{\frac{\Delta }{r_2}-\frac{\Delta }{r_1}}$
$\Rightarrow \frac{r_2}{r_2-r_3}=\frac{r_2}{r_1-r_2}$
$\Rightarrow r_1-r_2=r_2-r_3$
$\Rightarrow 2r_2=r_1+r_3$
$\therefore r_1,r_2,r_3\text{ are in AP. }$