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 $△GBC$ and $△GEF$ 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 $△BCM\sim △BDA$. 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 $△ABC$ and $△CDA$: $$\frac{AB}{BC}=\frac{AI}{CI}=\frac{AD}{CD}.$$ 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.