AustriaMO2013

Let $F$ be the diametrically opposite point to $D$ on $k$. Since $F$ is a point on the bisector of $BC$, we have $FB=FC$, and therefore $\angle BEF=\angle CEF$.



We see that $EF$ is the internal bisector of $\angle BEC$, and since $ED\perp EF$, $ED=AD$ is the external bisector. If we name $BC\cap EF=H$, we see that $A$ and $H$ are harmonic with respect to $B$ and $C$, since they are the points of intersection of perpendicular bisectors $EA$ and $EF$ with $BC$. Since $A$ is independent of the choice of $k$, we see that $H$ must be as well, and since $BC:CE=BH:CH$, the ratio is independent of the choice of $k$, as claimed. □