Question 84
Let $ABC$ be an acute triangle, $\Gamma$ its circumcircle and $O$ its circumcenter. Let $F$ be the point on $AC$ such that $\angle COF = \angle ACB$, where $F$ and $B$ lie on opposite sides of $CO$. The line $FO$ cuts $BC$ at $G$. The line parallel to $BC$ through $A$ intersects $\Gamma$ again at $M$. The lines $CO$ and $MG$ meet at $K$. Show that the circumcircles of the triangles $BGK$ and $AOK$ meet on $AB$.