Given triangle $ABC$ in which $\angle CAB= 30^{\circ}$ and $\angle ACB=60^{\circ}$. On the ray $AB$ a point $D$ is chosen, and on the ray $CB$ a point $E$ is chosen such that $\angle BDE=60^{\circ}$. Lines $AC$ and $DE$ intersect at $F$. Prove that the circumcircle of $AEF$ passes through a fixed point, which is different from $A$ and does not depend on $D$.