Consider that $AD=EF,\angle ODA=\angle OEF,OD=OE$ Hence $\triangle ODA\cong \triangle OEF$ Which means that $\triangle ODE\sim\triangle OAF$ So $\angle ODF=\angle OAF\Rightarrow O,F,A,D$ are cyclic

Dear Mathlinkers, another point of view by considerind the pivot theorem... Sncerely Jean-Louis

Since $OB=OD$, $OF=AC=AB$ and $\angle ABO=90-\angle DEB=90-\angle DBE=\angle ODF$ we have $\triangle OBA\cong \triangle ODF$, which gives that $\angle DFO=\angle BAO=\angle DAO$, which completes the solution.

Since $AB=AC, DE//AC$ we have $BD=DE$,then $OD$ is angle bisector of $\angle BDE$ from parallelogram ADFC we have $DF=AC=AB=AD+DB$ which means $DF=AD+2OD\cos\angle ODB$ then $DF\sin\angle ODB=AD\sin\angle ODB+OD\sin2\angle ODB$ So $DF\sin\angle ADO=AD\sin\angle ODE+OD\sin2\angle ADE$ According to Ptolemy's theorem, $O,F,A$ and $D$ lie on a circle.