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

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

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

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