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For any Δ ABC show that 2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2

Considering L.H.S :

\[2\text{ }\left( bc\text{ }cos\text{ }A\text{ }+\text{ }ca\text{ }cos\text{ }B\text{ }+\text{ }ab\text{ }cos\text{ }C \right)\]

\[2ca\text{ }cos\text{ }B,\text{ }2ab\text{ }cos\text{ }C\text{ }and\text{ }2cb\text{ }cos\text{ }A,\]are present in L.H.S. which can be obtained from cosine formulae.

Using cosine formula:

\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]

\[2bc\text{ }cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})\text{ }\ldots \text{ }\left( i \right)\]

Or,

\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]

\[2ac\text{ }cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})\ldots \text{ }\left( ii \right)\]

Or,

\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]

\[2ab\text{ }cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})\text{ }\ldots \text{ }\left( iii \right)\]

Adding equation (i), (ii) and (ii) we get,

\[2bc\text{ }cos\text{ }A\text{ }+\text{ }2ac\text{ }cos\text{ }B\text{ }+\text{ }2ab\text{ }cos\text{ }C\]

\[=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})\text{ }+\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})\text{ }+\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})\]

Simplifying,

\[=\text{ }{{c}^{2}}~+\text{ }{{b}^{2}}~+\text{ }{{a}^{2}}\]

So,

\[2\text{ }\left( bc\text{ }cos\text{ }A\text{ }+\text{ }ac\text{ }cos\text{ }B\text{ }+\text{ }ab\text{ }cos\text{ }C \right)\]

\[=\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }{{c}^{2}}\]

Hence proved.