$\measuredangle (EO,BC)=\measuredangle EOC+\measuredangle OCB=\measuredangle EAC+(90^\circ-\measuredangle BAC)=90^\circ$ Therefore $EO\perp BC$, similarly $DO\perp AB$, so $O$ is the orthocenter of $BED$, i.e. $(BED)$ and $(EOD)$ are of the equal size, as desired.

Let the circumcenters of (AOC) and (BDE) be M and N. We can see BE=EC because ∠OBC=∠OCB and as points A, O, C, E are concyclic, we have ∠OCE=∠OAB=∠OBE, therefore, ∠EBC=∠ECB. So, it suffices to show that triangles NBE and MEC are congruent. ∠NBE=∠BEN=90°−∠BDE= ∠EDC−90°=∠MEC=∠MCE, so triangles NBE and MEC are congruent and ME=MC=NB=NE, hence, the triangles BDE and AOC have circumradiuses of same length.