Q.In $\triangle ABC$, prove that $(b+c-a)\tan\dfrac{A}{2} = (c+a-b)\tan\dfrac{B}{2} = (a+b-c)\tan\dfrac{C}{2}$.

$b+c-a=2(s-a)$, and $\tan\dfrac A2=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$, so $(b+c-a)\tan\dfrac A2=2(s-a)\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}=2\sqrt{\dfrac{(s-a)^2(s-b)(s-c)}{s(s-a)}}=2\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}=\dfrac{2\sqrt{s(s-a)(s-b)(s-c)}}{s}=\dfrac{2\,A(\triangle ABC)}{s}$ (using Heron's Formula). This final expression is symmetric in $a,b,c$ (it does not privilege any one vertex), so by the identical computation, $(c+a-b)\tan\dfrac B2$ and $(a+b-c …

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