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

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(C) Given $(a-b)(s-c)=(b-c)(s-a)$.Using the relation $s-a = \frac{\Delta}{r_1}$,$s-b = \frac{\Delta}{r_2}$,and $s-c = \frac{\Delta}{r_3}$,we have $a-b

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$2 r_2$

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(C) Given $(a-b)(s-c)=(b-c)(s-a)$.Using the relation $s-a = \frac{\Delta}{r_1}$,$s-b = \frac{\Delta}{r_2}$,and $s-c = \frac{\Delta}{r_3}$,we have $a-b = (s-b)-(s-a) = \Delta(\frac{1}{r_2} - \frac{1}{r_1}) = \Delta \frac{r_1-r_2}{r_1 r_2}$.Substituting these into the given equation:$\Delta \frac{r_1-r_2}{r_1 r_2} \cdot \frac{\Delta}{r_3} = \Delta \frac{r_2-r_3}{r_2 r_3} \cdot \frac{\Delta}{r_1}$.Canceling $\frac{\Delta^2}{r_1 r_2 r_3}$ from both sides,we get:$r_1-r_2 = r_2-r_3$.Therefore,$r_1+r_3 = 2r_2$.

In a triangle $ABC$,if $(a-b)(s-c)=(b-c)(s-a)$,then $r_1+r_3=$

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