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.
 A: 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 area $I$.
Take any deformation $A^*B^*C^*D^*$, where the arcs and the areas are moved under translation and rotation. 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.
