A triangle ABC is inscribed in a circle $\omega$. $BB_1 $ bisects $\angle ABC$ (and so $M$ is the midpoint of the arc $AC$ ($B \notin AC$, where $AC$ is the arc)). $B_1K \perp BC$ ($K\in\omega$). $BL \perp AK$ ($L \in AC$). Prove that $K,L$ and $M$ are in a straight line.
I've used the letter D to represent those perpendicular angles in the picture (just so that they're easily seen). If you want to hear what I've tried, well, I can see that $\angle LBC = \angle AKB_1$, because $H_1H_2KB$ can be inscribed in another circle. I can't really tell anything more that could possibly be useful for solving this. I'm out of ideas and don't know where to start. Thanks.