Brahmagupta's formula (area of a cyclic quadrilateral)

Dependencies: Heron's formula, Ptolemy's theorem, Opposite angles of a cyclic quadrilateral are supplementary.

Statement

Let $ABCD$ be a cyclic quadrilateral with side lengths in order

and semiperimeter

Then the area $S$ of the cyclic quadrilateral $ABCD$ is uniquely determined by the four sides:

This formula is exactly parallel in form to Heron's formula (Heron's formula: $S = \sqrt{s(s-a)(s-b)(s-c)}$) — substituting the "triangle special case $d \to 0$" (i.e. vertex $D$ merging with $A$) into Brahmagupta's formula, the extra factor $(s - d) = s$ recovers Heron's formula. So Heron's formula is the special case of Brahmagupta's formula on a triangle, and the two form a "triangle ↔ cyclic quadrilateral" duality.