Balkan Mathematical Olympiad

Let $\Gamma$ be the image of $\gamma$ under the homothety of centre $A$ and factor $2$. Clearly, $\Gamma$ is also tangent to $AE$ at $A$, and the conclusion is equivalent to $\Gamma$ passing through $O$, which is the same as $AE$ being tangent to the circle $ACO$.

Alternatively, but equivalently, this amounts to $\angle OAE=\angle OCA$. Write $\angle OAE=90^{\circ }-\angle EBA$ and $\angle OCA=\angle OCB-\angle ACB=\angle OCB-\angle CBA=\angle OCB-\angle EBA$, to infer that the equality of the two angles is equivalent to $C$ being the midpoint of the segment $BE$.

To prove the latter, it is sufficient to show that the triangles $ABE$ and $DCB$ are similar, for then $BE/BC=AB/CD=AC/CD=2$, which implies that $C$ is indeed the midpoint of the segment $BE$.

Finally, to prove the above similarity, write $\angle EBA=\angle CBA=\angle ACB=\angle BCD$ and $\angle BDC=\angle BAD+\angle DBA=\angle BAD+\angle DAE=\angle BAE$. This completes the proof.