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:

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

So we get:

$\mathrm{\Delta }$$=$${\displaystyle \frac{1}{4}}\sqrt{-a^4-b^4-c^4+2b^2{c}^2+2a^2{b}^2+2a^2{c}^2}$

$=$${\displaystyle \frac{1}{4}}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$

$=$$\sqrt{s(s-a)(s-b)(s-c)}.$

This is Heron’s formula. $\mathrm{\square }$