Q:

Triangle area proof

Proof that the area of a triangle with sides $2a,2b,2c$ equals $$\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$
I have no idea how to tackle this, as I'm not very familiar with proofs. Can anybody gives me some clues/insights on how to tackle this problem?

A:

Heron's Formula can be used here.  If a triangle has sides $a,b,c$, then using the semiperimeter $s=\frac{a+b+c}{2}$, then area of a triangle is 
$$\sqrt{s(s-a)(s-b)(s-c)}$$
Now, instead of $a,b,c$, substitute in $2a, 2b, $and $2c$.  This also means your semiperimeter is 
$$s=\frac{2a+2b+2c}{2}=a+b+c$$
Thus, 
$$A=\sqrt{s(s-2a)(s-2b)(s-2c)}$$
$$=\sqrt{(a+b+c)(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)}$$
$$=\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$

