Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic

Theorem
Let $Q$ be a quadrilateral whose sides are $a$, $b$, $c$ and $d$.

Let $\AA$ be the area of $Q$.

Then:
 * $Q$ is a cyclic quadrilateral


 * $\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.
 * $\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.

Proof
From Bretschneider's Formula:
 * $\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$

where $s$ is the semiperimeter of $Q$.

Hence $\AA$ is greatest exactly when:
 * $\map {\cos^2} {\dfrac {\alpha + \gamma} 2} = 0$

That is, when:
 * $\map \cos {\dfrac {\alpha + \gamma} 2} = 0$

When this happens:


 * $\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$

which is Brahmagupta's Formula.

That is, $\AA$ is the area of a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$.

This happens exactly when:


 * $\dfrac {\alpha + \gamma} 2 = \dfrac \pi 2$

That is, when:


 * $\alpha + \gamma = \pi = 180 \degrees$

Hence the result from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles.