AUT_ABooklet_2023

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.