Belarus2022

Let $G$ be the point on the ray $BC$ such that $BG=3BC$. We will show that $G$ lies on the circumcircle of the triangle $AEF$. Angles $\angle ECF$ and $\angle ACG$ are equal $120^{\circ }$ as they are adjacent to $\angle ACB=60^{\circ }$. Since $\angle BED=30^{\circ }$ and $\angle FCE=120^{\circ }$, $\angle CFE=30^{\circ }$. Hence the triangle $FCE$ is isosceles with equal sides $CF=CE$. The leg $CB$ opposite to the angle $30^{\circ }$ is equal to half of the hypotenuse $AC$, so $AC=2CB=BG-CB=CG$. Hence the triangle $GCA$ is isosceles, $CG=CA$ and $\angle GAC=\angle AGC=30^{\circ }$. Note that the common bisector $\ell$ of the angles $ACG$ and $ECF$ is also the common perpendicular bisector of the segments $AG$ and $EF$. Hence the triangles $AEF$ and $GFE$ are symmetrical to each other with respect to $\ell$, and the centers of their circumcircles lie on line of symmetry. So they have a common circumscribed circle, i. e. point $G$ lies on the circumscribed circle of triangle $AEF$.