Given triangle $ABC$, $AL$ bisects angle $\angle BAC$ with $L$ on side $BC$. Lines $LR$ and $LS$ are parallel to $BA$ and $CA$ respectively, $R$ on side $AC$ and$ S$ on side $AB$, respectively. Through point $B$ draw a perpendicular on $AL$, intersecting $LR$ at $M$. If point $D$ is the midpoint of $BC$, prove that that the three points $A, M, D$ lie on a straight line.