Q:

In $\triangle ABC$, prove $4\left(\,bc \cos^2 \frac{A}2 + ca \cos^2 \frac{B}{2} + ab \cos^2 \frac{C}{2}\,\right) = (a + b + c)^2$

In any $\triangle ABC$, with sides $a$, $b$, $c$ opposite respective angles $A$, $B$, $C$, prove that:
  $$4\left(\,bc \cos^2\frac{A}{2} + ca \cos^2 \frac{B}{2} + ab \cos^2 \frac{C}{2}\,\right) = (a + b + c)^2$$

I am currently in class 11th and I am having trouble solving this. I know this is not a homework solving website, but I have no other option. Sorry. I will never again post a homework question.

A:

Using $$2\cos^2 \theta-1 =\cos 2\theta,$$L.H.S becomes:
$$2(bc(1+\cos A)+ca(1+\cos B)+ab(1+\cos C))=$$
$$2(bc+ca+ab)+2(bc\cos A +ca\cos B+ab\cos C)$$
Now use the cosine law
$$\cos A= \frac{b^2+c^2-a^2}{2bc} ,\cos B=\cdots$$ to get
$$2(bc+ca+ab)+(a^2+b^2+c^2)=$$
$$(a+b+c)^2$$

