Finding the quadrilateral with a fixed side and perimeter that has a maximum area

Let $ABCD$ be the quadrilateral ($AB=BC=CD$) in its "equal angles" configuration (i.e. $BCparallel AD$) and consider a small displacement $AB'C'D$ from that configuration $(AB=AB'=B'C'=C'D)$: we want to prove that the area of $ABCD$ is greater than the area of $AB'C'D$. Let's suppose $B'$ and $C'$ are the result of a counterclockwise rotation of $B$ and $C$ about $A$ and $D$ respectively, as in figure below.

As $BB'to 0$, segments $BB'$ and $CC'$ form the same angle with line $BC$; moreover, their projections on $BC$ must be equal (to ensure $B'C'=BC$). This implies that $BB'=CC'$ and that triangle $ABB'$ is congruent to triangle $DCC'$.

Hence the difference of the areas of quadrilaterals $ABCD$ and $AB'C'D$ is the same as the difference of the areas of triangles $MBB'$ and $MCC'$, where $M$ is the intersection point of $BC$ and $B'C'$. But these triangles have the same altitude upon bases $BM$ and $CM$, while $CM>BM$: it follows that the area of $MCC'$ is greater than the area of $MBB'$ and thus the area of $ABCD$ is greater than the area of $AB'C'D$, as it was to be proved.

EDIT.

The argument given above only proves that the "equal angles" configuration is a local maximum. To prove it is a global maximum one could use Bretschneider's formula for the area of a quadrilateral, which in our case gives: $$ area = {1over4}sqrt{(a+d)^3(3a-d)-16a^3dcos^2{alpha+gammaover2}}, $$ where $a=AB=BC=CD$, $d=AD$, $alpha=angle DAB$ and $gamma=angle BCD$.

This formula obviously gives the maximum area when $cos{alpha+gammaover2}=0$, that is when $alpha+gamma=pi$. But that condition implies the quadrilateral is cyclic and is then equivalent to $angle ABC=angle BCD$.