Area of Triangle in Terms of Circumradius

Theorem
Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.

Then the area $\mathcal A$ of $\triangle ABC$ is given by:
 * $\mathcal A = \dfrac {a b c} {4 R}$

where $R$ is the circumradius of $\triangle ABC$.

Proof

 * CircumradiusLengthProof.png

Let $O$ be the circumcenter of $\triangle ABC$.

Let $\mathcal A$ be the area of $\triangle ABC$.

Let a perpendicular be dropped from $C$ to $AB$ at $E$.

Let $h := CE$.

Then:

Let a diameter $CD$ of the circumcircle be passed through $O$.

By definition of circumradius, $CD = 2 R$.

By Thales' Theorem, $\angle CAD$ is a right angle.

By Angles on Equal Arcs are Equal, $\angle ADC = \angle ABC$.

It follows from Sum of Angles of Triangle Equals Two Right Angles that $\angle ACD = \angle ECB$.

Thus by Equiangular Triangles are Similar $\triangle DAC$ and $\triangle BEC$ are similar.

So: