For any triangle $A B C$ prove that $2 (b c cos A + c a cos B + a b cos C) = (a^{2} + b^{2} + c^{2})$ .

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Solution

Now for any $Δ A B C$ ,
$2 (b c cos A + c a cos B + a b cos C)$
$= 2 (b c cos A + c a cos B + a b cos C)$
$= (b^{2} + c^{2} - a^{2}) + (a^{2} + c^{2} - a^{2}) + (a^{2} + b^{2} - c^{2})$ [ Since $cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ and similarly for $cos B, cos C$ ]
$= (a^{2} + b^{2} + c^{2})$

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