Since $\angle FDE=\angle B+\angle C=180-\angle A$, $AFDE$ is cyclic. $$\measuredangle FED =\measuredangle FAD=\measuredangle BAD=\measuredangle BDE=\measuredangle CDE \implies EF\parallel BC$$So, the homothety at $A$ sending $\triangle ABC$ to $\triangle AFE$ also sends the circumcenter of $\triangle ABC$ to the circumcenter of $\triangle AFE$. In particular, $A$ and the circumcenters of the triangles $ABC$ and $DEF$ are collinear.

Easy angle chasing shows $AFDE$ cyclic.Then,draw the tangent at $A$ to $(ABC)$.Simple angle chasing shows it is also tangent to $(AFDE)$ .So,it means the two circles are homothetic from $A$ .