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

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(A) Given: $(a-b)(s-c)=(b-c)(s-a)$Since $a = (s-b) + (s-c)$,$b = (s-a) + (s-c)$,and $c = (s-a) + (s-b)$,we have $(a-b) = (s-b) - (s-a)$ and $(b-c) = (

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Arithmetic progression

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(A) Given: $(a-b)(s-c)=(b-c)(s-a)$Since $a = (s-b) + (s-c)$,$b = (s-a) + (s-c)$,and $c = (s-a) + (s-b)$,we have $(a-b) = (s-b) - (s-a)$ and $(b-c) = (s-c) - (s-b)$.Substituting these into the equation:$((s-b)-(s-a))(s-c) = ((s-c)-(s-b))(s-a)$$(s-b)(s-c) - (s-a)(s-c) = (s-c)(s-a) - (s-b)(s-a)$Rearranging terms:$2(s-a)(s-c) = (s-b)(s-c) + (s-b)(s-a)$Dividing both sides by $(s-a)(s-b)(s-c)$:$\frac{2}{s-b} = \frac{1}{s-a} + \frac{1}{s-c}$Multiplying by $\Delta$ (area of the triangle):$\frac{2\Delta}{s-b} = \frac{\Delta}{s-a} + \frac{\Delta}{s-c}$Since $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we get:$2r_2 = r_1 + r_3$Thus,$r_1, r_2, r_3$ are in Arithmetic Progression.

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

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