KöMaL 2016 B4810

Two angles of triangle $ABC$ are $BAC\sphericalangle=15^{\circ}$ and $ABC\sphericalangle=30^{\circ}$. The perpendicular drawn to side $AC$ at point $C$ intersects line segment $AB$ at $D$, and the perpendicular bisector of $AB$ intersects line $CD$ at $E$. Extend line segment $AB$ beyond $B$ by the length of line segment $BC$ to get $G$. Prove that the points $B$, $G$, $E$, $C$ lie on a circle of diameter $\sqrt{2} \cdot AB$.