In the acute-angled triangle $ABC$, let $AD$, $D \in BC$ be the $A$-angle bisector. The perpenducular to $BC$ through $D$ and the perpendicular to $AD$ through $A$ meet at $I$. The circle with center $I$ and radius $ID$, intersects $AB$ and $AC$ at $F$ and $E$ respectively. On the arc $FE$, which does not contain $A$, of the circumcircle of $AFE$, consider a point $X$, such that $\frac{XF}{XE}=\frac{AF}{AE}$. Prove that the circumcircles of triangles $AFE$ and $BXC$ are tangent.