In triangle ABC,frac{r_1-r}{a}+frac{r_2-r}{b} | vedclass

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Substituting these into t

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$\frac{r_1+r_2+r_3}{s}$

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Substituting these into the expression:$\frac{r_1-r}{a} = \frac{1}{a} \left( \frac{\Delta}{s-a} - \frac{\Delta}{s} \right) = \frac{1}{a} \left( \frac{\Delta s - \Delta(s-a)}{s(s-a)} \right) = \frac{\Delta a}{a s(s-a)} = \frac{\Delta}{s(s-a)} = \frac{r_1}{s}$.Similarly,$\frac{r_2-r}{b} = \frac{r_2}{s}$ and $\frac{r_3-r}{c} = \frac{r_3}{s}$.Adding these together:$\frac{r_1}{s} + \frac{r_2}{s} + \frac{r_3}{s} = \frac{r_1+r_2+r_3}{s}$.

In $\triangle ABC$,$\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c} =$

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