China Southeastern Mathematical Olympiad

First, if $A$, $B$, $C$, $D$ are concyclic, by $AB=DE$ and $BC=EA$ we have $\angle BAC=\angle EDA$, $\angle ACB=\angle DAE$, so $\angle ABC=\angle DEA$, which means that $AC=AD$.



Second, if $AC=AD$, let $O$ be the center of the circle ($B$, $C$, $D$, $E$ are on the circle). Then $O$ is on the perpendicular bisector ($AH$) of $CD$. Let $F$ be the symmetric point of $B$ by the line $AH$. Then $F$ is on the circle $O$. $AB\ne EA$, so $DE\ne DF$, and thus $E$, $F$ are not the same points. Now one can see that $△AFD\cong △ABC$, with $AB=DE$, $BC=EA$, and one can get $△AED\cong △CBA$, and so $△AED\cong △DFA$, which indicates $\angle AED=\angle DFA$, and therefore $A$, $E$, $F$, $D$ is concyclic. This means that $A$ is on the circle $O$, and $B$, $C$, $D$ are concyclic.