In the circumcircle of triangle $ABC$ we have $\angle COA = 2\angle CBA$. In the circumcircle of $ADC$ we therefore have $\angle CDA = \angle COA = 2\angle CBA$. The angle $\angle CDA$ is an external angle in triangle $ABD$, and we therefore obtain $\angle CBA + \angle BAD = \angle CDA = 2\angle CBA$, and thus $\angle BAD = \angle CBA$. In the circumcircle of $AOC$ we obtain $\angle BAD = \angle EAD$ on the chord $ED$. The angles $\angle CBA = \angle DBE$ are equal in the circumcircle of triangle $BDE$ on the same chord $ED$. Since the chords and subtended angles are equal in both circles, they must have the same radii, as claimed.