Let the intersection of $AC$ and $BD$ be $O$. Since $\angle AOD=\angle AFD=90^\circ$, $A,D,F,O$ are concyclic. Then, $\angle AFO = \angle ADO = 40/2=20^\circ$. Thus, $\angle OFE=70^\circ = \angle ACB$. Thus, $O,F,C,E$ are also concyclic. Then, $\angle CFE=\angle COE=\angle OCE=70^\circ$. Therefore, $\angle DFC=180^\circ-\angle EFC=110^\circ$.