In triangle ABC, with usual notations, if Del | vedclass

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(D) We know that the semi-perimeter $s = \frac{a+b+c}{2}$, so $2s = a+b+c$.Thus, $b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$.The formula for $	an(\frac

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$4\Delta$

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(D) We know that the semi-perimeter $s = \frac{a+b+c}{2}$, so $2s = a+b+c$.Thus, $b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$.The formula for $	an(\frac{A}{2})$ is $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.Substituting these into the expression: $2s \cdot 2(s-a) \cdot \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.$= 4s(s-a) \cdot \frac{\sqrt{(s-b)(s-c)}}{\sqrt{s(s-a)}}$.$= 4 \sqrt{s(s-a)(s-b)(s-c)}$.By Heron's formula, $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.Therefore, the expression equals $4\Delta$.

In $\triangle ABC$, with usual notations, if $\Delta$ denotes the area of triangle $ABC$, then the value of $2s(b + c - a) \tan(\frac{A}{2})$ is equal to ...

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