Cono Sur Mathematical Olympiad

Since $D$ and $E$ are midpoints we know that $DE\parallel AB$, and by definition $PR\parallel BC$, hence $PRDB$ is a parallelogram. Analogously, $QSDC$ is a parallelogram. Thus $PR=BD=DC=SQ$.



Since $AD$ is a diameter, we know that $AB\perp PD$ and $AC\perp QD$. Then, because of the parallel lines, $PD\perp DR$ and $QD\perp DS$. Hence $PR$ and $SQ$, which have the same length, are diameters of the circumcircles of $DPR$ and $DQS$ respectively.

In the cyclic quadrilateral $PXRD$, we have $\angle DPX=90^{\circ }$, so $DX$ is a diameter of the circumcircle of $DPR$. Likewise, $DY$ is a diameter of the circumcircle of $DQS$.

Since $DX$ is a diameter, we have that $XZ\perp ZD$, and since $DY$ is a diameter we have that $YZ\perp ZD$. Therefore $X,Z,Y$ are collinear. But we also know that both dashed circles have the same diameter, so $DX=DY$. Therefore $DZ$ is an altitude of the isosceles triangle $DXY$; this implies that $Z$ is the midpoint of $XY$, and we are done.