1. Define the points and circumcenters:
Let $O_1$ and $O_2$ be the circumcenters of the triangles $ABE$ and $ADE$, respectively. We need to show that the points $A, E, O_1,$ and $O_2$ form a square.

2. Circumradius equality:
Let $R_1$ and $R_2$ be the lengths of the circumradii of the triangles $ABE$ and $ADE$, respectively. Using the Law of Sines in triangles $ABE$ and $ADE$, we have:
$$2R_1 = \frac{AB}{\sin \angle AEB} \quad \text{and} \quad 2R_2 = \frac{AD}{\sin \angle AED}$$
Since $AB = AD$ (both are sides of the square $ABCD$) and $\angle AEB = \angle AED$ (as $E$ lies on the diagonal $BD$), it follows that:
$$2R_1 = 2R_2 \implies R_1 = R_2$$
Therefore, the circumradii are equal, and we have:
$$AO_1 = O_1E = R_1 = R_2 = EO_2 = O_2A$$

3. Angle calculation:
Next, we need to show that $\angle AO_1E = 90^\circ$. The angle subtended by the arc $AE$ in the circumcircle of $\triangle ABE$ is twice the angle at $B$:
$$\angle AO_1E = 2 \cdot \angle ABE$$
Since $\angle ABE = 45^\circ$ (as $\triangle ABE$ is a right triangle with $\angle ABE$ being half of $90^\circ$), we have:
$$\angle AO_1E = 2 \cdot 45^\circ = 90^\circ$$

4. Conclusion:
From the equal circumradii and the right angle, we conclude that $AO_1EO_2$ is a square. The sides are equal, and the angles are right angles.

$\blacksquare$