proof of Heron’s formula

Let $\alpha$ be the angle between the sides $b$ and $c$ , then we get from the cosines law:

\cos\alpha=\frac{b^{2}+c^{2}-a^{2}}{2bc}.

Using the equation

\sin\alpha=\sqrt{1-\cos^{2}\alpha}

we get:

\sin\alpha=\frac{\sqrt{-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2a^{2}b^{2}+2a^{2}c^{2}}% }{2bc}.

Now we know:

\Delta=\frac{1}{2}bc\sin\alpha.

So we get:

$\displaystyle\Delta$	$\displaystyle=$	$\displaystyle\frac{1}{4}\sqrt{-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2a^{2}b^{2}+2a^{2% }c^{2}}$
	$\displaystyle=$	$\displaystyle\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
	$\displaystyle=$	$\displaystyle\sqrt{s(s-a)(s-b)(s-c)}.$

This is Heron’s formula. $\Box$