Question:
In ∆ABC, prove the following:
$2(b c \cos A+c a \cos B+a b \cos C)=a^{2}+b^{2}+c^{2}$
Solution:
$\mathrm{LHS}=2(b c \cos A+c a \cos B+a b \cos C)$
On using the cosine law, we get:
$\mathrm{LHS}=2\left[b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)+c a\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)+a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\right]$
$=b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}$
$=a^{2}+b^{2}+c^{2}=\mathrm{RHS}$
Hence proved.