In triangle $A B C$
$2 (b c cos A + c a cos B + a b cos C) = a^{2} + b^{2} + c^{2}$

True

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False

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Solution

The correct option is A True
Using cosine rule $cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

Now, $2 (b c cos A + c a cos B + a b cos C)$

$= 2 (b c \times \frac{b^{2} + c^{2} - a^{2}}{2 b c} + c a \times \frac{c^{2} + a^{2} - b^{2}}{2 c a} + a b \times \frac{a^{2} + b^{2} - c^{2}}{2 a b})$

$= 2 (\frac{b^{2} + c^{2} - a^{2}}{2} + \frac{c^{2} + a^{2} - b^{2}}{2} + \frac{a^{2} + b^{2} - c^{2}}{2})$

$= 2 (\frac{c^{2} + a^{2} + b^{2}}{2})$

$= a^{2} + b^{2} + c^{2}$

Hence the given statement is true.

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