Consider homothethy centered at $E$ with scale $2$. You'll see that circumcenter-formed quadrilateral is cyclic iff lines perpendicular to $XE$ passing through $X$, where $X\in\lbrace A,B,C,D\rbrace$ form a cyclic quadrilateral. It's equivalent with $$(180^\circ-\angle AED)+(180^\circ-\angle BEC)=180^\circ$$Let $P$ be possibly other point than $A$ lying on line $AB$ such that $AE=PE$. Then $\angle EPB=\angle 180^\circ-\angle EAB=\angle 180^\circ-\angle ECB$ so points $B,C,E,P$ are concyclic. Take point $F$ such that $CF||DE\wedge AE||BF$ - that's just a translation of triangle $AED$ to side $BC$. We have $BF=AE=PE$ and $\measuredangle FBP=\measuredangle FBA=\measuredangle EAB=\measuredangle EAP=\measuredangle APE=\measuredangle BPE$ so points $B,P,E,F$ form an isosceles trapezoid. Thus they are concyclic. As a conclusion we have $\angle BEC=180^\circ-\angle CFB=180^\circ-\angle DEA$