Question
With usual notations prove that $2(b c \cos A+a c \cos B+a b \cos C)=a^2+b^2+c^2$.
$=2 b c\left(\frac{b^2+c^2-a^2}{2 b c}\right)+2 a c\left(\frac{c^2+a^2-b^2}{2 c a}\right)+2 a b\left(\frac{a^2+b^2-c^2}{2 a b}\right) \ldots($ By cosine rule $]$
$=b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2=a^2+b^2+c^2=$ RHS.
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