Heron formula for area of a triangle

Submitted by Structure on Sun, 07/11/2010 - 07:32.

Let ABC be a triangle, and a=length(BC), b=length(CA) and c=length(AB).Let $p=\frac{a+b+c}{2}$
Then area S of the triangle is

$S=\sqrt{p(p-a)(p-b)(p-c)}$

We know that

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$2\sin^2\frac{A}{2}=1-\cos A=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}=\frac{(a-b+c)(a+b-c)}{2bc}=2\frac{(p-b)(p-c)}{bc}$

Similar

$2\cos^2\frac{A}{2}=1+\cos A=1+\frac{b^2+c^2-a^2}{2bc}=\frac{(b+c)^2-a^2}{2bc}=\frac{(a+b+c)(-a+b+c)}{2bc}=2\frac{p(p-a)}{bc}$

so we have

$\sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}$

and

$\cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}$

Now we know

$S=\frac{bc\sin A}{2}=bc \sin\frac{A}{2}\cos\frac{A}{2}=bc\sqrt{\frac{(p-b)(p-c)}{bc}}\sqrt{\frac{p(p-a)}{bc}}=\sqrt{p(p-a)(p-b)(p-c)}$