Let the line through $E$ perpendicular to $BC$ intersect $CD$ at $P$ and $BC$ at $Q.$ it is trivial to verify $P$ lies on $(ACE).$ $P$ is also the orthocenter of $BCE$ so $(ECQD)$ and $(BDPQ)$ are cyclic.
Let $(EQB)$ intersect the line through $B$ parallel to $AC$ at $R.$ Let $(ECQD)$ intersect the line through $D$ perpendicular to $DO$ at $J.$ By radical axis it suffices to show $J$ lies on $(DBR).$
Notice that $\angle EQB = \angle ERB = 90^\circ,$ and $\angle EBR = \angle BAC = \angle EBQ,$ thus $R$ is the reflection of $Q$ across $EB$ and $(DBR)$ is the reflection of $(BDPQ)$ across $AB,$ thus $\angle DRB = \angle BEC$ and it suffices to show $\angle CJD = \angle BJD.$
Let the circle with diameter $AB$ intersect $(ACE)$ at $K.$ Since $KA$ and $JD$ are both perpendicular to $DO,$ they are parallel. $\angle KCE = \angle KAE = \angle JDB = \angle JDE$ so $K$ lies on $JC.$ Notice that $\angle AKB$ is right and it easily follows that $DJ$ is the perpendicular bisector of $BK$ implying the result. $\square$