Prove that (r + $r_{1}$ ) tan[(B - C)/2] + (r + $r_{2}$ ) tan[(C - A)/2] + (r + $r_{3}$ ) tan[(A - B)/2] = 0.

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Solution

We know that
$t a n \frac{B - C}{2} = \frac{b - c}{b + c} c o t \frac{A}{2}$
$∴ (r + r_{1}) t a n \frac{B - C}{2} = (\frac{S}{s} + \frac{S}{S - a}) \frac{b - c}{b + c} \sqrt{(\frac{s (s - a)}{(s - b) (s - c)})}$
$= \frac{S (2 s - a)}{s (s - a)} \frac{b - c}{b + c} \frac{\sqrt{s (s - a)}}{\sqrt{(s - b) (s - c)}}$
$= \frac{s (b + c)}{\sqrt{(s - a) (s - b) (s - c)}} \frac{b - c}{b + c}$
$= \frac{S}{S} (b - c) = (b - c)$
$∴$ $L H S = b - c + c - a + a - b = 0$

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