Solution:

From the conditions it follows that $\overline{FE}=\overline{AE}-\overline{AF}=\overline{AB}-(\overline{AB}-\overline{BC})=\overline{BC}$.
Therefore $BCEF$ is a parallelogram, whose two diagonals $CF$ and $BE$ bisect each other at $M$.

Because of the point symmetry about $M$, $\text{CBE= AEB}$ holds, and since $ABE$ is an isosceles triangle, $\text{AEB= EBA}$ holds. Thus $EB$ is the angle bisector of $\text{CBA}$. It follows that $\text{FDC= CBA=2 AEB}$, so that it only remains to show that $\text{AEB= ADB}$.

Let $D^{\prime}$ be the reflection point of $D$ about $M$. Since $DM \perp MO$, we then have

$\overline{OD^{\prime}}=\overline{OD}=\overline{OB}=\overline{OC}$ and $D^{\prime}BCD$ is a cyclic quadrilateral. Therefore $\text{BDC= BD C}$, and because $BCDEFD^{\prime}$ is even a hexagon that is point-symmetric about $M$, it further follows that $\text{EDF= BD C= BDC}$. Thus $\text{EDB= FDC}$, while on the other hand $\text{BAE=180 -2 AEB=180 - FDC}$ holds. Hence $ABDE$ is a cyclic quadrilateral, and the claim follows from the inscribed angle theorem.