2012 BMOSL G7 wrote:
$ABCD$ is a cyclic quadrilateral. The lines $AD$ and $BC$ meet at X, and the lines $AB$ and $CD$ meet at $Y$ . The line joining the midpoints $M$ and $N$ of the diagonals $AC$ and $BD$, respectively, meets the internal bisector of angle $AXB$ at $P$ and the external bisector of angle $BYC$ at $Q$. Prove that $PXQY$ is a rectangle
Claim- The Intersection of $\angle DXC$ and $\angle BYC$ lies on $MN$
Let the angle bisector of $\angle DXC$ intersect $MN$ at $P$. Now notice that $\triangle XND\sim\triangle XMC$ and $\triangle YNB\sim\triangle YMC$. So, $\frac{XN}{XM}=\frac{ND}{MC}=\frac{NB}{MC}=\frac{YN}{YM}$. Clearly as $XP$ is the Bisector of $\angle NXM$. Hence, $\frac{XN}{XM}=\frac{NP}{MP}=\frac{YN}{YM}$. Hence, $\angle NYP=\angle MYP$. Also $\angle BYN=\angle MYC$. Hence, $YP$ is the angle Bisector of $\angle BYC$.
Also simple angle chasing shows $\angle XPY=90^\circ$. Also $MN$ is the Gauss-Bondenmiller Line of $ABCD$. Hence, $MN$ bisects $XY$. Also $\angle QYP=90^\circ$ and we proved that $\angle XPY=90^\circ$. Hence, $PXQY$ is a rectangle. $\blacksquare$