In a triangle ABC, if (a-b)(s-c)=(b-c)(s-a), | vedclass

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(A) Given: $(a-b)(s-c)=(b-c)(s-a)$ $\Rightarrow \frac{s-c}{b-c}=\frac{s-a}{a-b}$ $\Rightarrow \frac{s-c}{(s-c)-(s-b)}=\frac{s-a}{(s-b)-(s-a)}$ Since $

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in arithmetic progression

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(A) Given: $(a-b)(s-c)=(b-c)(s-a)$ $\Rightarrow \frac{s-c}{b-c}=\frac{s-a}{a-b}$ $\Rightarrow \frac{s-c}{(s-c)-(s-b)}=\frac{s-a}{(s-b)-(s-a)}$ Since $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$, where $\Delta$ is the area of the triangle. $\Rightarrow \frac{\frac{\Delta}{r_3}}{\frac{\Delta}{r_3}-\frac{\Delta}{r_2}}=\frac{\frac{\Delta}{r_1}}{\frac{\Delta}{r_2}-\frac{\Delta}{r_1}}$ $\Rightarrow \frac{r_2}{r_2-r_3}=\frac{r_2}{r_1-r_2}$ $\Rightarrow r_1-r_2=r_2-r_3$ $\Rightarrow r_1+r_3=2r_2$ Therefore, $r_1, r_2, r_3$ are in arithmetic progression.

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

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