51st Ukrainian National Mathematical Olympiad, 3rd Round

Consider the diagonals of $MNCD$, $O$ is the point of its intersection, which is a center of the circle with diameter $DN$ (fig. 17). We have the following equalities: $$\angle NXC=\angle NDC=\angle MCD=\angle MXD.\text{ Hence,}$$ $$\angle MXC=\angle MXD+\angle DXC=\angle CXN+\angle DXC=\angle DXN=90^{\circ },$$