Question

In a $\triangle ABC$, 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 } \triangle A B C $
$ 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 2 r_2=r_1+r_3$
$ \therefore r_1, r_2, r_3 \text { are in AP. }$