A geometry question on finding the area of cyclic quadrilaterals The circumcircle of a cyclic quadrilateral $ABCD$ has radius $2$. 
$AC, BD$ meet at $E$ such that $AE = EC$. If $AB^2 = 2\cdot AE^2$ and $BD^2 = 12$, what is the area of the quadrilateral?

Let $O$ be the circumcentre of $ABCD$ and $R$ the circumradius. Since $BD=R\sqrt{3}$, $\widehat{DAB}$ and $\widehat{DCB}$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, in some order. Let $AE=x$. We have $AC=2x$ and $AB=x\sqrt{2}$, so:
$$ \arcsin\frac{2x}{4}=\arcsin\frac{x\sqrt{2}}{4}+\arcsin\frac{BC}{4} $$
and:
$$ BC = 2x\sqrt{1-\frac{x^2}{8}}-x\sqrt{2-\frac{x^2}{2}}. $$
In the same way you may compute $AD$ and $CD$ in terms of $x$. By Ptolemy's theorem we have $AB\cdot CD+BC\cdot AD = AC\cdot BD = 4x\sqrt{3}$; moreover, $BE+EC=2\sqrt{3}$ and $BE\cdot ED=x^2$.
Along the previous lines, you may find every possible value for $x$, then the length of every segment in the configuration. Notice that the area of $ABCD$ is given by:
$$ \frac{1}{2}\left(BC\cdot CD\sin\widehat{BCD}+AB\cdot AD\sin\widehat{BAD}\right)=\frac{\sqrt{3}}{4}\left(BC\cdot CD+DA\cdot AB\right).\tag{1}$$
We may also use the following approach. We take $BD$ as the side of an equilateral triangle inscribed in a circle $\Gamma$, and $A\in\Gamma$. We take $A'$ as the symmetric of $A$ with respect to $BD$, then $C_1,C_2$ as the intersections between $\Gamma$ and the parallel to $BD$ through $A'$. In such a way, if we define $E_i=AC_i\cap BD$, we have $AE_i=E_i C_i$. Then we take $F$ on the segment $AC_1$ such that $AF=AB\sqrt{2}$.

It is not difficult to check that $F$ may lie on the $BD$ line only if $C_1=C_2$. That gives:
$$ AB=2R\sin\frac{\pi}{12},\quad BC=CD=2R\sin\frac{\pi}{6},\quad AD=2R\sin\frac{5\pi}{12} $$
hence by $(1)$:
$$ [ABCD]=R^2\sqrt{3}\left(\sin^2\frac{\pi}{6}+\sin\frac{\pi}{12}\sin\frac{5\pi}{12}\right)=R^2\frac{\sqrt{3}}{2}=\color{red}{2\sqrt{3}}.\tag{2}$$
