Prove $r_{1} (r_{2} + r_{3}) / a = r_{2} (r_{3} + r_{1}) / b = r_{3} (r_{1} + r_{2}) / c$

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Solution

$\frac{r_{1} (r_{1} + r_{3})}{a} = \frac{\frac{S}{s - a} [\frac{S}{s - b} + \frac{S}{s - c}]}{a}$
$= \frac{S}{a (s - a)} [S (\frac{1}{s - b} + \frac{1}{s - c})]$
$= \frac{S^{2}}{a (s - a)} (\frac{s - c + s - b}{(s - b) (s - c)})$
$= \frac{S^{2} (2 S - (b + c))}{a (s - a) (s - b) (s - c)}$
$= \frac{S^{2} a s}{a S^{2}} = s$
$\frac{r_{2} (r_{1} + r_{3})}{b} = \frac{S}{b - (s - b)} [S (\frac{1}{s - a} + \frac{1}{s - c})]$
$= \frac{S^{2}}{b (s - b)} (\frac{2 s - (a + c)}{(s - a) (s - c)})$
$= \frac{S^{2} b}{b (s - b) (s - a) (s - c)} \frac{S^{2} b}{b S^{2} / s} = s$
$\frac{r_{3} (r_{1} + r_{2})}{c} = \frac{S}{c (s - c)} [S (\frac{1}{(s - a)} + \frac{1}{(s - b)})]$
$= \frac{S^{2}}{c (s - c)} \frac{(2 s - (a + b))}{(s - a) (s - b)}$
$= \frac{S^{2} c}{c S^{2} / s} = s$
$\Rightarrow \frac{r_{1} (r_{2} + r_{3})}{a} = \frac{r_{2} (r_{1} + r_{3})}{b} = \frac{r_{3} (r_{1} + r_{2})}{c} = s$ proved.

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