I am not sure if it is correct. Let $\angle ABE=a,\angle BCA=b,\angle CDB=c,\angle DEC=d,\angle EAD=e$. Then $a\leq b\leq c\leq d\leq e$. Since $\angle BAC\leq \angle BDC$, the point $A$ lies outside or on the circumcircle of $\triangle BCD$, so $\angle DAC\leq \angle DBC=c$. Similarly, $A$ lies inside or on the circumcircle of $\triangle AED$, so $\angle DAC\geq\angle CED=d$. Thus we have $c\geq \angle CAD\geq d$, but $c\leq d$, so we must have $c=d=\angle CAD$. Then $BCDA,EDCA$ are cyclic and $b=c=d=e$ follows. From here, it can be easily proven one way or another that all the angles are equal.