Let $M$ be the midpoint of $CD$ and $X = BC \cap ED$. Since $\angle ABX = \angle AEX = 90^\circ$,
AX is the diameter of the circumcircle of $\triangle ABX$, thus $ACXD$ is a parallelogram.

Now, it is sufficient to show that $[OAB] = [OAE]$ where $[OAB]$ denotes the area of $\triangle OAB$.
Let $C', D'$ be the midpoint of $AC$ and $AD$. It is easy to see that $[OAB] = [D'AB] = \frac{1}{2}[ACD]$ and therefore $[OAE] = [C'AE] = [C'AD] = \frac{1}{2}[ACD]$. We are done.

As in Solution 1. $ACXD$ is a parallelogram and $\angle DAB = 90^\circ$.
Let $N$ be the midpoint of $BE$. It is enough to show that $\angle NAB = \angle OAB$. By the cyclic of $ABXE$, we have
$$\angle ABE = \angle AXE = \angle XAC$$
and $\angle BEA = \angle CXA$. Therefore, $\triangle ABE \simeq \triangle CAX$, and $N$ corresponds to $M$ under this similarity. In particular, $\angle NAB = \angle ACM$. We have
$$\angle OAB = 90^\circ - \angle DAO = \angle ACM = \angle NAB.$$