Problem 1381

Let $M$ be the intersection of the angle bisector of $\angle B D C$ with the circumcircle of $\triangle A B C$. Since $D$ lies on the short arc $A C$, $M$ lies on the arc $B C$ where $A$ does not lie. We have $\angle B D M=\angle M D C$ because $D M$ is the angle bisector of $\angle B D C$, so arcs $B M$ and $C M$ are of equal length. This implies that the position of $M$ does not depend on the position of $D$.

Let $S$ be the intersection of $D E$ and the circumcircle of $\triangle A B C$. We will prove that $S$ does not depend on the position of $D$. Since $S$ and $M$ lie on the circumcircle of $\triangle A B C$, we have $\angle A M D=\angle A S D=\angle A S E$. Since $E$ is the reflection of $A$ in $D M$, we now see that $\angle A M E=2 \angle A M D=2 \angle A S E$. Consider the circle with center $M$ passing through $A$. Due to the reflection, $|M A|=|M E|$, so this circle also passes through $E$. The central angle theorem now tells us that from $\angle A M E=2 \angle A S E$ it follows that $S$ also lies on this circle. We see that $S$ is the second intersection of the circumcircle of $\triangle A B C$ and the circle with center $M$ passing through $A$. This fixes $S$, independently of the position of $D$. Since $D E$ passes through $S$, $S$ is the desired point.