Solution:
Recall that, in every triangle, the altitude and the diameter of the circumcircle drawn from the same vertex are isogonal. The proof offers no difficulty, being a simple angle chasing around the circumcircle of the triangle.
Let $O$ be the circumcenter of the triangle $ABC$. From the above, one has $\angle OAE = 90^{\circ} - B$. On the other hand $\angle DEA = B$, for $BCDE$ is cyclic. Thus $AO \perp DE$, implying that in the triangle $ADE$ cevians $AO$ and $AO_{2}$ are isogonal. So, since $AO$ is a radius of the circumcircle of triangle $ABC$, one obtains that $AO_{2}$ is an altitude in this triangle.

Moreover, since $OO_{1}$ is the perpendicular bisector of the line segment $BC$, one has $OO_{1} \perp BC$, and furthermore $AO_{2} \parallel OO_{1}$.

Chord $DE$ is common to $\omega_{1}$ and $\omega_{2}$, hence $O_{1}O_{2} \perp DE$. It follows that $AO \parallel O_{1}O_{2}$, so $AOO_{1}O_{2}$ is a parallelogram. The conclusion is now obvious.