Proof of necessicity: Suppose $A,B,C,D$ are concyclic. $AB=DE$ and $BC=EA$ implies that $ABDE$ and $ABCE$ are trapezoid. So $AC=BE=AD$. Proof of sufficiency: Suppose $AC=AD$. Triangles $ABC$ and $DEA$ are congruent. Let $F$ be the point such that $CDEF$ is an isosceles trapezoid ($EF\parallel CD$). Hence $B,C,D,E,F$ are concyclic. We also have $\angle ACF=\angle DAE=\angle BAC$ and $CF=DE=AB$, hence $ABFC$ is an isosceles trapezoid, which implies that $A$ also lies on the circumcircle of $BCDEF$. Thus $A,B,C,D$ are concyclic.