A diameter $AK$ is drawn for the circumscribed circle $\omega$ of an acute-angled triangle $ABC$, an arbitrary point $M$ is chosen on the segment $BC$, the straight line $AM$ intersects $\omega$ at point $Q$. The foot of the perpendicular drawn from $M$ on $AK$ is $D$, the tangent drawn to the circle $\omega$ through the point $Q$, intersects the straight line $MD$ at $P$. A point $L$ (different from $Q$) is chosen on $\omega$ such that $PL$ is tangent to $\omega$. Prove that points $L$, $M$ and $K$ lie on the same line.