In ∆ABC, prove the following:
$2 (b c \cos A + c a \cos B + a b \cos C) = a^{2} + b^{2} + c^{2}$

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Solution

LHS = $2 (b c \cos A + c a \cos B + a b \cos C)$

On using the cosine law, we get:
$LHS = 2 [b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) + c a (\frac{a^{2} + c^{2} - b^{2}}{2 a c}) + a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b})]$
$= b^{2} + c^{2} - a^{2} + a^{2} + c^{2} - b^{2} + a^{2} + b^{2} - c^{2} = a^{2} + b^{2} + c^{2} = RHS$

Hence proved.

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