The South African Mathematical Olympiad Third Round

Let $X$ be the intersection of $SM$ with the circumcircle. Note that $\angle MXA=\angle SXA=\angle SBA=\angle SBC+\angle CBA$. Since $AS$ bisects the angle $\angle BAC$, we have $\angle SBC=\angle SAC=\angle BAS$. It follows that $$\angle MXA=\angle BAS+\angle CBA=\angle BAD+\angle DBA=180^{\{}\circ \}-\angle ADB=180^{\{}\circ \}-\angle ADM,$$ so $AXMD$ is a cyclic quadrilateral. Consequently, $\angle SXD=\angle MXD=\angle MAD=\angle DAN=\angle SAQ=\angle SXQ$, which means that $X,D$ and $Q$ lie on a common straight line. Hence $X$ is also the intersection of $SM$ and $QD$, which completes the proof.