Let $M$ be the midpoint of $AE$ (so $OM \perp AE$), and let $O'$ be the reflection of $O$ over $M$. Then I will show $O'$ is the circumcenter of $\triangle ADE$. Since $AB \perp AE,$ $AB \parallel OO'$, and since $\angle O'AE = \angle OAE = \angle OEA$, $O'A \parallel BE$. Thus, $ABOO'$ is a parallelogram. Thus, a translation by the vector $\overrightarrow{BA}$ maps $BC$ to $AD$ and $O$ to $O'$. Since $O$ lies on the perpendicular bisector of $BC$, then $O'$ must lie on the perpendicular bisector of $AD$. Also, it is clear that $O'$ lies on the perpendicular bisector of $AE$. Thus, $O'$ is the circumcenter of $\triangle ADE$, so since $O'A=OA$, the two circles have the same radius.

$E$ - othocenter of $\triangle ACD$.