Let $D, E$ be the feet of the perpendiculars form $A, C$ onto $BC, AB$ respectively. Notice that $DE\| BK$, and also $KBDE$ is cyclic, because the points lie on the circle with diameter $BH$, hence $KBDE$ is an isosceles trapezoid $\implies BD=KE$ and since $L$ is the center of $AEDC$ we have $DL=EL$, now easy angle chasing leads to $\angle BDL=\angle KEL$ and $\triangle BDL \cong \triangle KEL$ so done $\blacksquare$

I had the same solution as Ricochet, but I've elaborated a little more on the details (such as why $DE \parallel BK$ and why $\angle BDL = \angle KEL$) in my Youtube video here: https://www.youtube.com/watch?v=f5BGt6FgkhA

Let $X$, $Y$ be the midpoint of $BH$ and $BK$. Let $D, E, F$ be the foot of the perpendiculars from the orthocenter to $BC, AB, AC$ respectively. $\angle BXY= 90- (\angle KBH)=90-(\angle KBC-\angle KBH)=90-(180-\angle A - (90- \angle C))=\angle A- \angle C$ $\angle HXL= \angle FXL = \angle FEL=\angle AEL- \angle AEF=\angle A - \angle AHF = \angle A - \angle C$ we considered that X, E, F, L are points on the nine-point circle and thus concyclic; AEHF and FHDC are cyclic Combining the two gives us $\angle BXY= \angle FXL \implies X, Y, L$ are collinear. $XY$ is the midline of triangle $BKH$ so it is perpendicular to $BX$ since $\angle BXH$ is a right triangle. Thus, $LY$ bisects and is perpendicular to $BK$. The conclusion follows.