In any ΔABC, prove that

Need to prove: $4\left(b c \cos ^{2} \frac{A}{2}+c a \cos ^{2} \frac{B}{2}+a b \cos ^{2} \frac{C}{2}\right)=(a+b+c)^{2}$

Right hand side

$=4\left(b c \cos ^{2} \frac{A}{2}+c a \cos ^{2} \frac{B}{2}+a b \cos ^{2} \frac{C}{2}\right)$

$=4\left(b c \frac{s(s-a)}{b c}+c a \frac{s(s-b)}{c a}+a b \frac{s(s-c)}{a b}\right)$, where s is half of perimeter of triangle.

$=4(s(s-a)+s(s-b)+s(s-c))$

$=4\left(3 s^{2}-s(a+b+c)\right)$

We know, $2 s=a+b+c$

So, $4\left(3\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\right)^{2}-\frac{(\mathrm{a}+\mathrm{b}+\mathrm{c})^{2}}{2}\right)$

$=4\left(3 \frac{(a+b+c)^{2}}{4}-\frac{(a+b+c)^{2}}{2}\right)$

$=4\left(\frac{3(a+b+c)^{2}-2(a+b+c)}{4}\right)$

$=3(a+b+c)^{2}-2(a+b+c)^{2}$

$=(a+b+c)^{2}$

$=$ Right hand side. [Proved]