The first thing I noticed was $\triangle CDM \cong \triangle BCN$ and we obtain $CM = BN$ and $\angle MCD = \angle NBC$. Now I am trying to show $\angle BFM = 90 ^\circ$ but I am having difficulties with this and don't know if it will help with the solution. Would appreciate help of any kind!
Edit: Thank you for your help! Now I see how $\angle BFM = 90^\circ$. $\angle BAM + \angle BFM = 180 ^\circ$, thus $ABFM$ is a cyclic quadrilateral. Does this help?