$AA', BB'$ and $CC'$ heights of an acute triangle $ABC$. The circle with center $B$ and radius $BB'$ intersects the line $A'C'$ in the points $K$ and $L$. Prove that the intersection point of lines $AK$ and $CL$ lies on the line $BO$, where $O -$ center of the circle circumscribed about the triangle $ABC$.
My wotk so far:
$H -$ orthocenter of triangle $ABC$ is the center of the inscribed circle of triangle $A'B'C'$
