Let $F$ be the second intersection of $\omega_1$ and $\omega_2$. Notice that
$$\angle BFO_2 = 180^\circ - \angle BAO_2 = \angle CAO_2 = \angle O_2CA$$
and since $AO_2 = EO_2$ that $\angle FBO_2 = \angle O_2BA$. It follows that $F$ is the reflection of $C$ over $BO_2$, thus it suffices to prove that $EF$ is parallel to $BO_2$, as then $BO_2$ is a midline of triangle $CEF$.

Notice that $\angle O_2DC = \angle DCO_2 = \angle EAD$. Hence by symmetry about $O_1O_2$, we have
$$\begin{align*} \angle FEO_2 &= \angle AED = 180^\circ - \angle EAD - \angle ADE \\ &= \angle ADO_2 - \angle CDO_2 = \angle ADC = \frac{1}{2} \angle AO_2C \end{align*}$$

Moreover
$$\begin{aligned} 90^\circ - \angle O_1O_2B &= \frac{1}{2}\angle BO_1O_2 = 180^\circ - \angle BAO_2 \\ &= \angle CAO_2 = 90^\circ - \frac{1}{2}\angle AO_2C \end{aligned}$$
so we conclude that $\angle FEO_2 = \angle EO_2B$ which proves that $EF$ and $BO_2$ are parallel.

Let $G$ be the point on $\omega_2$ diametrically opposite $C$. As in the first solution, it suffices to prove that $EG$ is parallel to $BO_2$, as then $BO_2$ is a midline of triangle $CEG$. Note first that
$$\angle GO_2E = 2\angle ECD = \angle DAE + \angle GAD = \angle GAE$$
so the points $A$, $E$, $G$, and $O_2$ lie on a circle. Since $GO_2 = AO_2$, we see that $O_2$ is the midpoint of the arc $AG$ of this circle. It follows then that
$$\angle GEO_2 = AGO_2 = \frac{1}{2}\angle AO_2C$$
and as we showed in the first solution this implies that $\angle GEO_2 = \angle EO_2B$, so lines $EG$ and $BO_2$ are parallel.