Let $G$ be the 2nd intersection of $DO$ with $(O)$ (reflection of D about O). By Pascal theorem for the degenerate cyclic hexagon $DCABBG,$ the intersections $P \equiv DC \cap BB,$ $E \equiv CA \cap GB$ and $O \equiv AB \cap GD$ are collinear. Since $\overline{ADF} \parallel \overline{BGE}$ and $\overline{OA}=-\overline{OB},$ then $AFBE$ is parallelogram with diagonal intersection $O,$ i.e. $O$ is midpoint of $\overline{EF}.$

According to the statement, E lies on line PO, so why do you use pascal theorem to prove P,E,O being collinear? otherwise, i don't understand how do you get $\overline{ADF}\parallel\overline{BGE}$?

I used Pascal theorem to prove that the points $B,G,E$ are collinear. It is a matter of denotation, intersections $P \equiv CD \cap BB,$ $E' \equiv CA \cap GB$ and $O \equiv AB \cap GD$ are collinear $\Longrightarrow$ $E \equiv E'.$

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Luis González wrote: I used Pascal theorem to prove that the points $B,G,E$ are collinear. It is a matter of denotation, intersections $P \equiv CD \cap BB,$ $E' \equiv CA \cap GB$ and $O \equiv AB \cap GD$ are collinear $\Longrightarrow$ $E \equiv E'.$ i got it now, thanks!