In $ riangle A B C$, if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1, r_2, r_3$ are in

Question

In $	riangle A B C$, if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1, r_2, r_3$ are in

Aicharya · Accepted Answer

Given: $(a-b)(s-c)=(b-c)(s-a)$ $\begin{aligned} & \Rightarrow[(s-a)-(s-b)](s-c)=[(s-b)-(s-c)](s-a) \ & \Rightarrow(s-a)(s-c)-(s-b)(s-c) \ & =(s-b)(s-a)-(s-c)(s-a)\end{aligned}$ Dividing by $(s-a)(s-b)(s-c)$ on both sides, we get $\begin{aligned} & \frac{1}{(s-b)}-\frac{1}{(s-a)}=\frac{1}{(s-c)}-\frac{1}{(s-b)} \ & \frac{2}{(s-b)}=\frac{1}{(s-a)}+\frac{1}{(s-c)} \ & \Rightarrow \frac{2 \Delta}{(s-b)}=\frac{\Delta}{(s-a)}+\frac{\Delta}{(s-c)} \ & \Rightarrow 2 r_2=r_1+r_3\end{aligned}$ So, $r_1, r_2, r_3$ are in arithmetic progression.

In $\triangle A B C$, if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1, r_2, r_3$ are in

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