On the hypotenuse $AB$ of a right-angled triangle $ABC$, point $R$ is chosen, on the cathetus $BC$ a point $T$, and on the segment $AT$ a point $S$ are chosen in such a way that the angles $\angle ART$ and $\angle ASC$ are right angles. Points $M$ and $N$ are the midpoints of the segments $CB$ and $CR$, respectively. Prove that points $M$, $T$, $S$, and $N$ lie on the same circle. Proposed by S. Berlov