Belarus competition problems

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 $AEF$.