IMO Team Selection Test 3, June 2020

Let $M$ be the intersection of the internal angular bisector of $\angle BDC$ with the circumcircle of $△ABC$. As $D$ lies on the short arc $AC$, we see that $M$ lies on the arc $BC$ not containing $A$. We have $\angle BDM=\angle MDC$ as $DM$ is the internal angular bisector of $\angle BDC$, so arcs $BM$ and $CM$ have equal lengths. Hence $M$ is independent of $D$.



Let $S$ be the intersection of $DE$ and the circumcircle of $△ABC$. We show that $S$ is independent of $D$. As $S$ and $M$ lie on the circumcircle of $△ABC$, we have $\angle AMD=\angle ASD=\angle ASE$. As $E$ is the reflection of $A$ in $DM$, we now see that $\angle AME=2\angle AMD=2\angle ASE$.

Consider the circle with centre $M$ passing through $A$. As $E$ is the reflection of $A$ in $DM$, we have $|MA|=|ME|$, so this circle also passes through $E$. By the inscribed angle theorem, from $\angle AME=2\angle ASE$ it follows that $S$ is also on this circle. Therefore $S$ is the second intersection point of the circumcircle of $△ABC$ and the circle with centre $M$ passing through $A$. This is a description of $S$ independent of $D$. As $DE$ passes through $S$, the point $S$ is the point required. $\square$