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

if and only if:

$\AA$ 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.

$\blacksquare$