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

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Solution

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

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