Suppose the symmetry point of $E$ about $AD$ is $B_1$, the symmetry point of $D$ about $EC$ is $B_2$. Then $A{B_1}DE$ and $EDC{B_2}$ are rhombuses, so $A{B_1}\parallel ED$ and $C{B_2}\parallel DE$. Let $AB=a$, because $AC<2a$, ${B_1} \ne {B_2}$.
Because $\angle D{B_1}S = \angle DES = \angle DCS$, so ${B_1}CDS$ on a circle, hence $\angle DC{B_1} = \angle AS{B_1} = 60^\circ $. Because $D{B_1} = a = DC$, $\Delta {B_1}CD$ is a equilateral triangle, so $C{B_1} = a = A{B_1}$. Also we have $A{B_2} = a = C{B_2}$.
Because there are only two points satisfy $CX = a = AX$, so $B = {B_1}$ or $B = {B_2}$, hence $AB\parallel ED$ or $CB\parallel DE$.
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