In triangle ABC,if r_1 = 2r_2 = 3r_3,then | vedclass

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(C) Given,in $	riangle ABC$,$r_1 = 2r_2 = 3r_3$. Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c

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$a+c=2b$

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(C) Given,in $	riangle ABC$,$r_1 = 2r_2 = 3r_3$. Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have: $\frac{\Delta}{s-a} = 2 \frac{\Delta}{s-b} = 3 \frac{\Delta}{s-c} = k$ (let). Then,$s-a = \frac{1}{k}$,$s-b = \frac{2}{k}$,and $s-c = \frac{3}{k}$. Adding these equations: $(s-a) + (s-c) = \frac{1}{k} + \frac{3}{k} = \frac{4}{k}$. Since $s-b = \frac{2}{k}$,we have $\frac{4}{k} = 2(s-b)$. Thus,$2s - a - c = 2s - 2b$,which simplifies to $a+c = 2b$.

In $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then

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