Let $P$ denote the reflection of $A$ over $O$,$E'=PC\cap \omega$ and $Q=EC\cap \odot{(BOD)}$. Since $OD\parallel PC$, we have $\angle{E'OD}=\angle{OE'P}=\angle{E'PO}=\angle{'EBA}$ which implies $E'=E$. By power of a point we have \begin{align*} CA\cdot{}CB &=CP\cdot{}CE \\ 2DC\cdot{}CB &=CP\cdot{CE} \\ 2QC\cdot{}CE &=CP\cdot{}CE \\ \implies{}PC &=2QC. \end{align*} Hence $Q$ is the midpoint of $PC$, thus $OQ\parallel AC$. Therefore $\angle{ECA}=\angle{CQO}=\angle{EFO}$, implying our desired result.