European Mathematical Cup

We prove the problem in reverse as this is much more natural in this problem.
We note that if $BCEF$ is a parallelogram then the diagonales are bisecting each other so the point $G \equiv BE \cap CF$ should be the midpoint of $CE$.
If $G$ is the midpoint of $CE$ then $\triangle GBC$ and $\triangle GEF$ are congruent as $CG = GF$ and $FE \parallel BC$ gives $\angle GEF = \angle GBC$ and $\angle GFE = \angle GCB$. Hence this implies $BG = GE$ and in particular $BCEF$ is a parallelogram as its diagonals bisect each other. Hence $G$ being midpoint of $CF$ is equivalent to our problem.
As $M$ is the midpoint of $AC$ by the midline theorem applied to triangle $ACF$ we have $G$ is the midpoint of $CG$ if and only if $MG \parallel AF$. Hence we only need to prove $BM \parallel AF$.
Now we further notice that, using $FD \parallel BC$, this is equivalent to $\angle AFD = \angle MBC$.
We further see that $\angle AFD = \angle ABD$ as they are angles over the same chord. So our claim is equivalent to $\angle ABD = \angle MBC$.
We add that here depending on the relative position of $F$ on the circles we might have $\pi - \angle AFD = \angle MBC$ but then $\pi - \angle AFD = \angle ABD$ so the final conclusion still holds.
We know that $\angle BDA = \angle BCM$ as they are angles over the same chord. Now this us that our claim is equivalent to the claim $\triangle BCM \sim \triangle BDA$.
The same angle equality shows that this is equivalent to $\frac{BC}{CM} = \frac{AD}{BD}$. Using the fact $M$ is the midpoint of $AC$ we have $CM = \frac{AC}{2}$ so our claim is equivalent to $2AD \cdot BC = BD \cdot AC$.
We further have by the angle bisector theorem applied to $\triangle ABC$ and $\triangle CDA$:
$$\frac{AB}{BC} = \frac{AI}{CI} = \frac{AD}{CD}.$$
So using this our claim is equivalent to $AB \cdot CD + AD \cdot BC = BD \cdot AC$ which we can recognise to the Ptolemy's theorem for cyclic quadrilaterals.