Question 172
Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$. Let $D$ be a point on segment $BC$ such that $\angle BAD = \angle DAC$, and let $M$ and $N$ be points on segments $BD$ and $CD$, respectively, such that $\angle MAD = \angle DAN$. Let $S, P$ and $Q$ (all different from $A$) be the intersections of the rays $AD, AM$ and $AN$ with $\Gamma$, respectively. Show that the intersection of $SM$ and $QD$ lies on $\Gamma$.