For any Δ ABC sho

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