$DB$ is a diameter of the bigger circle whose center is $A$.
$I$ and $C$ are the midpoints of the circular arcs $\widehat{DB}$
Point $E$ is a random point in line $DB$ and $F$ is the contact point between the bigger and smaller circle, the latter being also tangent to $DB$ on $E$.
Point $H$ is the meeting of $FB$ and the perpendicular to $DB$ through $E$.
Show that $D,H$ and $I$ are always collinear (as $E$ moves through $DB$)
Let $M$ be center of the other circle. $|FM|=|ME|$ and $|FA|=|AC|$ and so, $FME \sim FAC$. Hence, $F, E, C$ are collinear. Since $C$ is mid-point of the arc $BD$, $\angle DFC = \angle BFC = 45^\circ$. $DEHF$ is a cyclic quadrilateral and $\angle EDH = \angle EFH = \angle CFB = 45^\circ$ and we are done!