Question

In ΔABC, show that $\sum(\frac{r_1}{(s - b)(s - c)} = \frac{3}{r}$.

Answer 1

We take

$\frac{r_{1} }{(s-b)(s-c)} \\\\=\frac{\frac{\Delta}{s-a} }{(s-b)(s-c)}\\ \\=\frac{\Delta}{(s-a)(s-b)(s-c)}\\\\=\frac{s.\Delta}{s(s-a)(s-b)(s-c)}\\\\=\frac{s.\Delta}{\Delta^{2} }\\\\\\=\frac{s}{\Delta }\\\\\\=\frac{s/s}{\Delta/s}\\\\\\=\frac{1}{r}\\\\Similarly$

$\frac{r_{2} }{(s-a)(s-b)} = \frac{1}{r} \\\\And\\\\\frac{r_{3} }{(s-b)(s-c)} = \frac{1}{r} \\\\So\\\Sigma\frac{r_{1} }{(s-b)(s-c)} = \frac{1}{r}+\frac{1}{r}+\frac{1}{r}=\frac{3}{r}$