Indija TS 2008

Observe that $NX$ is the polar of $Y$, $EF$ is the polar of $A$ and $BC$ is the polar of $D$. Since $A$, $Y$, $D$ are collinear, it follows that $NX$, $EF$, $BC$ are concurrent. Let the point of concurrency be $D^{\prime }$. Let $S$ be the point of intersection of $EF$ and $AD$. Since \{$E,S,F,D^{\prime }$\} form a harmonic range, \{AE, AD, AB, AD'\} is a harmonic pencil. It follows that \{D', B, D, C\} is a harmonic range. We also observe that $\angle DN{D}^{\prime }=90^{\circ }$. Therefore $ND$ bisects $\angle BNC$. This implies that $D$ is the mid-point of the minor arc $PQ$ of $\Gamma$, where $P$, $Q$ are the points of intersection of $NB$, $NC$ with $\Gamma$. Hence $PQ$ is parallel to $BC$. Now $$\angle YNQ=\angle NPQ=\angle NBC.$$ It follows that $YN$ is tangent to the circum-circle of the triangle $BNC$.