MEMO 2012 I3

In a given trapezium $ABCD$ with $AB$ parallel to $CD$ and $AB > CD$ , the line $BD$ bisects the angle $\angle ADC$ . The line through $C$ parallel to $AD$ meets the segments $BD$ and $AB$ in $E$ and $F$ , respectively. Let $O$ be the circumcenter of the triangle $BEF$ . Suppose that $\angle ACO = 60^{\circ}$ . Prove the equality $$CF = AF + FO .$$