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Question
In ∆ABC, prove the following:
\[2 \left( bc \cos A + ca \cos B + ab \cos C \right) = a^2 + b^2 + c^2\]
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Solution
LHS = \[2 \left( bc \cos A + ca \cos B + ab \cos C \right)\]
On using the cosine law, we get:
\[LHS = 2\left[ bc\left( \frac{b^2 + c^2 - a^2}{2bc} \right) + ca\left( \frac{a^2 + c^2 - b^2}{2ac} \right) + ab\left( \frac{a^2 + b^2 - c^2}{2ab} \right) \right]\]
\[= b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2 \]
\[ = a^2 + b^2 + c^2 = RHS\]
Hence proved.
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