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In $\triangle \mathrm{ABC}$, with usual notations, $2 \mathrm{ab} \sin \frac{1}{2}(\mathrm{~A}+\mathrm{B}-\mathrm{C})=$
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The correct answer is:
$a^2+b^2-c^2$
$\begin{aligned} & 2 a b \sin \frac{1}{2}(A+B-C) \\ & =2 a b \sin \frac{1}{2}[(\pi-C)-C]=2 a b \sin \left(\frac{\pi-2 C}{2}\right) \\ & =2 a b \sin \left(\frac{\pi}{2}-C\right)=2 a b \cos C \\ & =2 a b\left(\frac{a^2+b^2-c^2}{2 a b}\right)=a^2+b^2-c^2\end{aligned}$
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