Problem

Source: Mexico 2023/5

Tags: geometry



Let $ABC$ be an acute triangle, $\Gamma$ is its circumcircle and $O$ is its circumcenter. Let $F$ be the point on $AC$ such that the $\angle COF=\angle ACB$, such that $F$ and $B$ lie in opposite sides with respect to $CO$. 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 $AB$.