# Geometric argument as to why the cyclic quadrilateral has the maximal area

I am looking for a purely geometric/intuitive argument as to why the cyclic quadrilateral has the maximal area among all quadrilaterals having the same side lengths.

I am aware of couple of proofs, which resort to some sort of algebra/ calculus/ trigonometric argument.

For instance, the Bretschneider's formula, gives the area of a quadrilateral as $$\sqrt{(s-a)(s-b)(s-c)(s-d) - abcd \cos^2(\theta/2)}$$ where $\theta$ is the sum of a pair of opposite angles of the quadrilateral. Hence, given $a,b,c,d$, the maximum is attained when $\theta = \pi$, which allows us to conclude that the quadrilateral has to be cyclic.

Another very similar argument, using calculus and trigonometry, is mentioned in this article.

However, I am not able to intuitively understand why among all quadrilateral with given sides $a,b,c,d$ the cyclic quadrilateral is the one that maximizes the area. I believe a geometric argument would provide me a good intuition in understand this non-trivial fact.

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My favorite interpretation comes from the following claim / 'obvious fact'

Claim: Amongst all smooth curves with a fixed perimeter $P$, the area $A$ satisfies the inequality $4 \pi A \leq P^2$. Equality holds when the curve is a circle.

Lemma: If the 4 given sides satisfy the cyclic inequalities $a < b+c+d$, then a quadrilateral with sides $a, b, c, d$ exists. Moreover, a cyclic quadrilateral with sides $a, b, c, d$ exists. A sketch of this is given at the end.

This Lemma merely serves to let us know when the question makes geometric sense. Now, let $ABCD$ be a cyclic quad with the desired side lengths, and consider it's circumcircle. Let it have area $X$. Fix the circular arcs along with the intermediate areas between the circular arcs and the quadrilateral. Let these arcs have perimeter $P$, and the intermediate area have total $I$.

Take any deformation $A^*B^*C^*D^*$, where the arcs and the areas are moved under translation. This gives a figure with perimeter $P$, and area $I+X^*$, where $X^*$ is the area of the deformed $A^*B^*C^*D^*$. By the initial claim, $I+X^* \leq I+X$. Hence we are done.$_ \square$

Sketch Proof of Lemma: The first part follows from the triangle inequality. The second part follows by taking a continuous deformation of the quadrilateral, showing that opposite angles must equal $180^\circ$ at some point in time.

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Wow. This is a very nice argument. Is this your argument? If not, could you let me know where you saw/read this? Thanks. – user17762 Dec 29 '12 at 2:39
This is not an original argument, and I've seen it in the past. Note that some issues have been swept under the rug by the initial claim. It might not be trivial to prove, though it's intuitively obvious. – Calvin Lin Dec 29 '12 at 2:54
Yes. True the iso-perimetric inequality, I believe, is a non-trivial inequality. But at-least your argument gave a good intuition into why the cyclic quadrilateral maximizes the area and what exactly this depends on. – user17762 Dec 29 '12 at 3:06
Note that this also works for the n-gon case, though the second part of the lemma might be tricky to justify easily. – Calvin Lin Dec 29 '12 at 8:20
Yes. True. Regarding the iso-perimetric inequality, I am only aware of algebraic/calculus proofs for it. I am not sure if there is a purely geometric argument, though I guess it is not that hard to come up with a semi-geometric argument for the iso-perimetric inequality. Anyways, your answer is a great one. Wish I could up-vote it more than once. – user17762 Dec 29 '12 at 8:28