Proof of Heron's Formula for the area of a triangle 
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.

Could you please provide the proof of this formula?
Thank you in advance.

A simple derivation exploits the cosine theorem. We have $\Delta=\frac{1}{2}ab\sin C$, hence
$$ 4\Delta^2 = a^2 b^2 \sin^2 C = a^2 b^2 (1-\cos C)(1+\cos C).\tag{1}$$
On the other hand, $ 2ab\cos C = a^2+b^2-c^2$, hence
$$ 2ab(1+\cos C) = (a+b)^2-c^2 = (a+b+c)(a+b-c), \tag{2}$$
$$ 2ab(1-\cos C) = c^2-(a-b)^2 = (a-b+c)(-a+b+c),\tag{3}$$
and by multiplying $(2)$ and $(3)$ and exploiting $(1)$
$$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)\tag{4} $$
which is equivalent to
$$ \Delta = \sqrt{s(s-a)(s-b)(s-c)}\tag{5} $$
as wanted.

Alternative derivation: by considering the circumcenter $O$ and its distances from the sides we have
$$2\Delta = R\sum_{cyc}a\cos A,\qquad 16\Delta^2 = \sum_{cyc}a^2\cdot 2bc(\cos A)=\sum_{cyc}a^2(b^2+c^2-a^2)\tag{6}$$
through $4R\Delta=abc$ and the cosine theorem. By rearranging the RHS of $(6)$
$$ 16\Delta^2 = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)\tag{7} $$
immediately follows.
