Denote $\angle DAE = \alpha$; then $\angle DOE = 2\alpha$ (Fig. 5). From the isosceles triangle $DOE$, we obtain $\angle OED = \frac{180^\circ - 2\alpha}{2} = 90^\circ - \alpha$. Since $DE$ and $AO$ are parallel, $\angle AOE = 90^\circ - \alpha$.

As the common tangent to $\omega_1$ and $\omega_2$ at point $A$ is perpendicular to the radius $AO$ of $\omega_1$, as well as to the radius of $\omega_2$, and $O$ lies on $\omega_2$, the line segment $AO$ must be a diameter of $\omega_2$. Thus $\angle OCA = 90^\circ$. As $OA = OD$, the line segment $OC$ is the altitude of the isosceles triangle $OAD$ drawn from its apex angle. This implies $\angle AOE = \angle DOE = 2\alpha$.
Hence $90^\circ - \alpha = 2\alpha$, implying $\alpha = 30^\circ$.

As $\angle ACO = \angle DCE$ and also $\angle AOC = \angle DEC$, the triangles $ACO$ and $DCE$ are similar.

Fig. 5

As the common tangent to $\omega_1$ and $\omega_2$ at point $A$ is perpendicular to the radius $AO$ of $\omega_1$, as well as to the radius of $\omega_2$, and $O$ lies on $\omega_2$, the line segment $AO$ must be a diameter of $\omega_2$. Thus $\angle OCA = 90^\circ$. As $OA = OD$, the line segment $OC$ is the altitude of the isosceles triangle $OAD$ drawn from its apex angle. This implies $AC = DC$. Hence the triangles $ACO$ and $DCE$ are equal and $AO = DE$.
A quadrilateral with a pair of equal parallel sides is a parallelogram. As $AO = DO$, the quadrilateral $AODE$ is a rhombus. The diagonals of a rhombus bisect the angles at their endpoints. Thus $\angle EAD = \frac{1}{2}\angle EAO$. As $OA = OE = OD$ and $AE = OD$, the triangle $AOE$ is equilateral. Hence the internal angles of $AODE$ are of size $60^\circ$ and $120^\circ$, implying $\angle EAD = 30^\circ$.