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

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.

## closed as off-topic by Daniel W. Farlow, JonMark Perry, Ben Sheller, Shailesh, choco_addictedMar 23 '16 at 3:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Daniel W. Farlow, JonMark Perry, Ben Sheller, Shailesh, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.

• Please change the title to something more meaningful. Also use MathJax wherever possible. And of course, welcome to Math.SE. You should show some effort on homework questions, and someone will gladly help – Shailesh Mar 22 '16 at 16:04
• furthermore define your variables – tired Mar 22 '16 at 16:05
• Reason for downvote? – hackware wright Mar 22 '16 at 16:08
• You are allowed to post homework questions. However, we just ask that you also show your thoughts on the problem. This may include a partial solutions or anything you have tried. – user320276 Mar 22 '16 at 16:21

## 1 Answer

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 $$2(bc+ca+ab)+(a^2+b^2+c^2)=$$ $$(a+b+c)^2$$