AustriaMO2013

We intersect the entire figure with the plane through the points $X$, $E$ and $F$. Since $M$ is on line $EF$, it is part of that plane. Also points $Y$ and $Z$ are part of that plane because they are on line $EX$. The intersection of the circumsphere $k$ and the plane results in a circle $k^{\prime }$, which also has $M$ as its center. The intersection of $ABCD$ with the plane is the perpendicular bisector of line $EF$, and $Y$ is on that bisector.

We denote $\angle ZEF=\alpha$. Since $ZM$ and $EM$ are both radii of $k$ (and $k^{\prime }$), the triangle $ZME$ is isosceles, therefore $\angle EZM=\angle ZEM=\alpha$. Since $Y$ is on the bisector of $EF$, also triangle $EYF$ is isosceles, therefore $\angle YFE=\angle YEF=\alpha$. Therefore the triangles $ZME$ and $EYF$ are similar, and their corresponding angles $\angle EMZ$ and $\angle EYF$ are identical. $\square$

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Alternative solution.

We intersect the figure with the plane $XEF$ as we did in solution 1. Since $\angle YMF=90^{\circ }$, and by Thales also $\angle FZY=90^{\circ }$, the quadrilateral $YMFZ$ has a circumcircle. Therefore $\angle YZM=\angle YFM$, and we again get that the triangles $ZME$ and $EYF$ are similar. $\square$