China Girls' Mathematical Olympiad

Let $M$ and $N$ be the feet of the perpendiculars from $O$ to the lines $DF$ and $DE$, respectively. Because $O$ is the circumcenter of the triangle $DEF$, the triangles $EOD$ and $ODF$ are both isosceles with $EO=DO=FO$. It follows that $$\angle EOF=\angle EOD+\angle DOF=2\angle NOD+2\angle DOM=2\angle NOM.$$ Because $\angle OND=\angle OMD=90^{\circ }$, the quadrilateral $OMDN$ is concyclic, from which it follows that $\angle NDM+\angle NOM=180^{\circ }$ or $\angle BDN=\angle NOM$. Because $ABED$ is concyclic, we have $\angle BAE=\angle BDE$. Combining the above equations together, one has $$\angle EOF=2\angle NOM=2\angle BDN=2\angle BDE=2\angle BAE.$$ Because $BC$ is a diameter of $\Gamma$, it follows that $$\angle EOF+\angle EAF=2\angle BAE+2\angle EAC=2\angle BAC=180^{\circ },$$ from which it follows that $AEOF$ is concyclic. Let $\omega$ denote the circumcircle of $AEOF$. Because $O$ lies on the perpendicular bisector of the segment $EF$, $O$ is the midpoint of the arc $EF$ (on $\omega$), implying that $AO$ bisects $\angle EAF$ and $A$, $C$, $O$ are collinear.