National Math Olympiad

The central angle is twice the inscribed angle, so $\angle CSD=2\angle CBD$. The triangle $CSD$ is isosceles with the apex at $S$, so $$\angle DCS=\frac{\pi -\angle CSD}{2}=\frac{\pi }{2}-\angle CBD.$$ Thus, $$\angle ACS=\angle ACD+\angle DCS=\angle CBD+\frac{\pi }{2}-\angle CBD=\frac{\pi }{2}.$$ Now, because $E$ is the midpoint of $BD$, the line $SE$ is perpendicular to the line $BD$ and $\angle SEA=\frac{\pi }{2}$. This implies $\angle ACS+\angle SEA=\pi$, so the points $A,E,S$ and $C$ are concyclic.