Heron's Formula/Proof 2

Theorem
Given a triangle $\triangle ABC$ with sides $a$, $b$, and $c$ opposite points $A$, $B$, and $C$, respectively.

Let $s$ be the semiperimeter, so $s = \dfrac{a + b + c} 2$.

Then the area $A$ of the triangle is given by the formula $A = \sqrt{s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}$.

Proof
A triangle can be considered as a cyclic quadrilateral one of whose sides has degenerated to zero.

From Brahmagupta's Formula, the perimeter of a cyclic quadrilateral is given by:
 * $\sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$

where $s$ is the semiperimeter:
 * $s = \dfrac{a + b + c + d} 2$

The result follows by letting $d$ tend to zero.