66th Czech and Slovak Mathematical Olympiad

Denote by $S$ the circumcenter of triangle $ABF$ and by $K$, $L$, $M$ the feet of perpendiculars from $S$ to lines $AB$, $BF$, $BE$. It's easy to check that angle $ABF$ is obtuse, point $S$ lies on the perpendicular bisector $CK$ of side $AB$ in the half-plane opposite to $CEB$, and that point $E$ lies inside segment $SL$ so it is the interior point of segment $BM$ (Fig. 3). As $M$ is the midpoint of the mentioned chord and $AC=BC$, it suffices to show $BM=BC$. Fig. 3

Denote by $\alpha$, $\beta$ the angles by bases of isosceles triangles $ABC$, $BFE$, respectively. Since $\angle BDF=\alpha$, triangle $DBF$ gives $$\angle CBM=180^{\{}\circ \}-2(\alpha +\beta ).$$ As $CE\parallel AB$, we further get $\angle BCE=\alpha$ and the clear similarity of right triangles $SEM\sim BEL$ yields $\angle ESM=\beta$. And since both $CE$ and $AB$ are perpendicular to $SK$, the quadrilateral $CEMS$ is cyclic implying $\angle ECM=\angle ESM=\beta$. Altogether, for $\angle BCM$ and $\angle BMC$ we get $$\angle BCM=\angle BCE+\angle ECM=\alpha +\beta$$ and from the previous equation we compute $$\angle BMC=180^{\{}\circ \}-\angle CBM-\angle BCM=2(\alpha +\beta )-(\alpha +\beta )=\alpha +\beta .$$ From $\angle BMC=\angle BCM$ we deduce $BM=BC$.