Let us denote the centers of the circumscribed circles of triangles ABE, BCE, CDE, ADE by $O_{AB}$, $O_{BC}$, $O_{CD}$, $O_{AD}$ respectively. Let $T_1, T_2$ be the points of tangency of one of the common tangents with the circumscribed circles of triangles ABE and CDE, respectively; denote by O and T the midpoints of segments $O_{AB}O_{CD}$ and $T_1T_2$, respectively (см. рис. 1). Then, in the right trapezoid $O_{AB}T_1T_2O_{CD}$, the line $OT$ is the midline, hence it is the perpendicular bisector of segment $T_1T_2$. Note that the circle $\omega$ is symmetric with respect to the line $O_{AB}O_{CD}$, on which the point $O$ also lies, meaning that $O$ is the center of $\omega$.

Рис. 1

Рис. 2

Similarly, we find that the midpoint of segment $O_{AD}O_{BC}$ is the center of $\gamma$. Therefore, the statement of the problem is equivalent to the fact that $O_{AB}O_{BC}O_{CD}O_{AD}$ is a parallelogram. To prove this, it suffices to note that $O_{AB}O_{BC}$ and $O_{CD}O_{AD}$ are the perpendicular bisectors of segments $EB$ and $ED$, respectively, hence $O_{AB}O_{BC} \parallel O_{CD}O_{AD}$; similarly, $O_{AB}O_{AD} \parallel O_{BC}O_{CD}$, from which the required conclusion follows.