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Let $F$ be the second intersection of ${\omega }_1$ and ${\omega }_2$. Notice that $$\angle BF{O}_2=180^{\circ }-\angle BA{O}_2=\angle CA{O}_2=\angle O_{2}CA$$ and since $A{O}_2=E{O}_2$ that $\angle FB{O}_2=\angle O_{2}BA$. It follows that $F$ is the reflection of $C$ over $B{O}_2$, thus it suffices to prove that $EF$ is parallel to $B{O}_2$, as then $B{O}_2$ is a midline of triangle $CEF$.

Notice that $\angle O_{2}DC=\angle DC{O}_2=\angle EAD$. Hence by symmetry about $O_1{O}_2$, we have $$\begin{array}{rl}\angle FE{O}_2& =\angle AED=180^{\circ }-\angle EAD-\angle ADE\\ & =\angle AD{O}_2-\angle CD{O}_2=\angle ADC=\frac{1}{2}\angle A{O}_{2}C\end{array}$$

Moreover $$\begin{array}{rl}90^{\circ }-\angle O_1{O}_{2}B& =\frac{1}{2}\angle B{O}_1{O}_2=180^{\circ }-\angle BA{O}_2\\ & =\angle CA{O}_2=90^{\circ }-\frac{1}{2}\angle A{O}_{2}C\end{array}$$ so we conclude that $\angle FE{O}_2=\angle E{O}_{2}B$ which proves that $EF$ and $B{O}_2$ are parallel.



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Alternative solution.

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 $B{O}_2$, as then $B{O}_2$ is a midline of triangle $CEG$. Note first that $$\angle G{O}_{2}E=2\angle ECD=\angle DAE+\angle GAD=\angle GAE$$ so the points $A$, $E$, $G$, and $O_2$ lie on a circle. Since $G{O}_2=A{O}_2$, we see that $O_2$ is the midpoint of the arc $AG$ of this circle. It follows then that $$\angle GE{O}_2=AG{O}_2=\frac{1}{2}\angle A{O}_{2}C$$ and as we showed in the first solution this implies that $\angle GE{O}_2=\angle E{O}_{2}B$, so lines $EG$ and $B{O}_2$ are parallel.