In $ riangle A B C, \frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}=$

Question

In $	riangle A B C, \frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}=$

Aicharya · Accepted Answer

In $	riangle A B C$, $ \begin{aligned} & \frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c} \ & =\frac{\frac{\Delta}{s-a}-\frac{\Delta}{s}}{a}+\frac{\frac{\Delta}{s-b}-\frac{\Delta}{s}}{b}+\frac{\frac{\Delta}{s-c}-\frac{\Delta}{s}}{c} \ & 	herefore \quad\left[r_1=\frac{\Delta}{s-a} r_2=\frac{\Delta}{s-b} r_3=\frac{\Delta}{s-c} 	ext { and } r=\frac{\Delta}{s}ight] \ & =\frac{\Delta(s-s+a)}{a s(s-a)}+\frac{\Delta(s-s+b)}{b s(s-b)}+\frac{\Delta(s-s+c)}{c s(s-a)} \ & =\frac{\Delta}{s-a} \cdot \frac{1}{s}+\frac{\Delta}{s-b} \cdot \frac{1}{s}+\frac{\Delta}{s-c} \cdot \frac{1}{s} \ & =\frac{r_1}{s}+\frac{r_2}{s}+\frac{r_3}{s}=\frac{r_1+r_2+r_3}{s} \ & \end{aligned} $

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

Solution