I found out this problem when I was fooling around with Geogebra. I've tried to solve it for some day but found nothing yet.
Let $ABC$ be an acute triangle inscribed the circumcenter $(O)$, three altitudes $AD, BE, CF$. $AD$ intersects $(O)$ at $K$. Let $I$ be the midpoint of $EF$, $BI$ intersects $(O)$ at $J$. Prove that $K, E, J$ are collinear.
My attempts: Let $L$ be the intersection of $AO$ and $(O)$. It's a well-known result that $\widehat{LAC}=\widehat{KAB}$.
Since $\widehat{KJB}= \widehat{KAB}$ we need to prove that $\widehat{LAC}=\widehat{KJB}$ or if we let $Q$ be the intersection of $AL$ and $BJ$, we need to prove that $AJQE$ is a cyclic quadrilateral.
We see that $\widehat{AJB} = \widehat{ACB}$, so we deduce to prove that $QE \parallel BC$.
And I stuck here.
Please help me to solve my problem.
Thanks for reading.