Given an acute $\Delta ABC$ $(AB<AC)$ with circumcenter $O$ and orthocenter $H$.
Line $AO$ intersects $BC$ at $D$.
Line segment $AC$ intersects the circumcircle of $\Delta ABD$ at point $E$.
Line segment $AB$ intersects the circumcircle of $\Delta ACD$ at point $F$.
Line segments $BE$ and $CF$ intersects at $K$.
Prove that $HK$ is parallel to $BC$.
I actually knew how to prove this and succesfully proved it, but I want to know if my proof is correct and if there is another better way, so I posted both the figure and my attempt as my own answer.