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 $$s = \frac{a + b + c}{2}$$.

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

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:
 * $$s = \frac{a + b + c + d}{2}$$

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