For any Δ ABC show that-

2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2


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Note: In any ΔABC we define ‘a’ as length of side opposite to A , ‘b’ as length of side opposite to B and ‘c’ as length of side opposite to C .


Key point to solve the problem:


Idea of cosine formula in ΔABC


Cos A =


Cos B =


Cos C =


As we have to prove:


2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2


As LHS contain 2ca cos B, 2ab cos C and 2cb cos A ,which can be obtained from cosine formulae.


From cosine formula we have:


Cos A =


2bc cos A = …..eqn 1


Cos C =


2ab cos C = …eqn 2


And, Cos B =


2ac cos B = ……eqn 3


Adding eqn 1,2 and 3:-


2bc cos A + 2ab cos C + 2ac cos B = +


2bc cos A + 2ab cos C + 2ac cos B =


2(bc cos A + ab cos C + ac cos B) = proved


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