We consider the inverion circle with centre $A$ and radius $\sqrt{AC\cdot AF}$. Therefore $C$ goes to $F$ and $F$ goes to $C$. What is more we know that if $D'$ is the inverse of $D$, then $AD\cdot AD'=AC\cdot AF$. However it is $AD=AC$ (tangents), so we deduce that $AF=AD'$. Let $E'$ be the inverse of $E$. We know that $AE\cdot AE'=AD\cdot AD'=AC\cdot AF$. We know that the circle $\omega$ becomes the line $CE'$ and that the circle $\Gamma$ becomes the circle that is tangent to $AF$ and $AD'$ at $F$ and $D'$ respetively. The intersection of these two is the point $B'$, the inverse of $B$. The circumcircle of triangle $ABC$ becomes the line $B'F$ and the circumcircle of triangle $DEB$ becomes the circumcircle of $D'E'B'$. The above circles are tangents if and only if $B'F$ is tangent to the circlumcircle of $D'E'B'$, that is to say if and only if $\widehat{FB'D'}=\widehat{D'E'B'}$. Let $\widehat{FB'D'}=x$. Its not hard to prove that $\widehat{FAD'}=2x-180^o$, so in order that $\widehat{D'E'B'}=x$, $\widehat{AE'C}$ should be $180^o-x$, so the triangle $AE'C$ must be isosceles, that is to say $AE'=AC$! To sum up the circumcircles of triangles $ABC$ and $DEB$ are tangent if and only if $AE'=AC$! We know that $AC=AD$, so $AE'=AD$. We also know that $AE\cdot AE'=AD\cdot AD'$, so $AE=AD'$. Finally we know that $AD\cdot AD'=AC\cdot AF\Leftrightarrow AD\cdot AE=AC\cdot AF$, that is to say that the points $C, D, F, E$ are cyclic!

this is just easy angle chasing(thanks to my teammate for giving me significant hint)