Let $M_{AF}$ be the midpoint of $AF$ and name the other $5$ midpoints similarly. By gliding principle, $M_{AF}M_{BC}M_{DE}$ is an equilateral triangle. Thus $M_{AF}$ is on the perpendicular bisector of $M_{BC}M_{DE}$. $M_{BC}M_{CD}=\frac{BD}{2}=M_{CD}M_{DE}$, so $M_{CD}$ is on the perpendicular bisector of $M_{BC}M_{DE}$. Thus $\overline{M_{AF}M_{CD}}$ passes through the circumcenter of $M_{AF}M_{BC}M_{DE}$, and by symmetry we're done.

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