Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic
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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
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.
$\blacksquare$