In any ` DeltaA B C ,2(bc cosA+ ca cosB+ab cosC) =`

Explanation

Given,LHS=` 2(bc cos A + ca cos B + ab cos C)`
Now, as LHS contain `(2ca cos B, 2ab cos C and 2cb cos A,)` terms so,by cosine formula we have `2bc cos A = (b^2 + c^2 – a^2) … (i) `

`2ac cos B = (a^2 + c^2 – b^2)… (ii)`

`2ab cos C = (a^2 + b^2 – c^2) … (iii) `

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

`2bc cos A + 2ac cos B + 2ab cos C = (b^2 + c^2 – a^2) + (a^2 + c^2 – b^2) + (a^2 + b^2 – c^2) `

= `c^2 + b^2 + a^2 `

Thus,`2(bc cos A + ac cos B + ab cos C) = a^2 + b^2 + c^2 `
Thus proved.