Let OC meet the circle (O) again at Q. CQ is a diameter of (O), hence $CD \perp DQ$ and $DQ \parallel AB.$ ADQBPC is a cyclic hexagon and by Pascal's theorem, the intersections $E \equiv AP \cap CQ,\ F \equiv BC \cap PD,\ G \equiv AB \cap DQ$ are collinear. Since $DP \parallel AB,$ their intersection G is at infinity parallel to AB. Hence the collinearity line $EF \parallel AB$ is a midline of the triangle $\triangle OBC.$

I think that I have an easier solution: By angles we have $CAO \sim CDB$ and $\angle CAE = \angle CDF$, thus$CF/BF = CE/OE$.

In my opinion , just notice that $\vartriangle AEO \sim \vartriangle DFB$ . Then it's not difficult to find that $C,E,F,P$ is concyclic. Then it's obviously that $EF//AB$ (by equal angles) ,so that we get that $F$ is the midpoint of $BC$ .

connect EF,∠OCB=∠OBC=∠APD,so,C,E,F,P are concyclic,therefore,∠CFE=∠CPA=∠CBA,so,EF∥AB,∵E is the midpoint of CO,∴F is the midpoint of CB,QED