$AY \parallel EB$ and $AYBE$ cyclic mean that $AYBE$ is a CyclicISLsosceles Trapezoid and since $BYCX$ is a parallelogram, we get that $ACXE$ is a CyclicISLsosceles Trapezoid so now, reflecting the line $\overline{ADX}$ over the perpendicular bisector of $AC$ we get that $C,D,E$ are colinear thus we are done

First note that $\measuredangle DCA =\measuredangle CAD = \measuredangle CAX = \measuredangle BXA $ and $\measuredangle CAD = \measuredangle DAB = \measuredangle XAB$. Thus, $A$ is the center of the spiral similarity mapping $DC$ to $BX$. In turn, it is also the center of the spiral similarity mapping $BD$ to $CX$ which gives us that $\measuredangle BDA = \measuredangle XCA = \measuredangle BYA$. Thus, $Y$ also lies on $(ABD)$. Now, since $A$ is the center of the spiral similarity mapping $BD$ to $CX$, the intersection of $(ABD)$ and $(ACX)$ besides $A$ must be the intersection of the lines $\overline{BX}$ and $\overline{CD}$. Let this point be $E'$. Then, $E' = (AYDB) \cap BX=E$. Thus, $E$ also lies on $(ACX)$ and $\overline{CD}$ which proves the required claim.

We shall prove an equivalent problem Quote: Let $\triangle ABC$ a triangle, and $D$ the intersection of the angle bisector of $\angle BAC$ and the perpendicular bisector of $AC$. the line parallel to $AC$ passing by the point $B$, intersect the line $AD$ at $X$. the line parallel to $CX$ passing by the point $B$, intersect $AC$ at $Y$. $E = (CD) \cap (BX) $ . Prove that $Y, A, B, E$ are concyclic.. Let <BAC =2w => <BAD = <DAC =< ACD = w because D lies on perp.bisector of AC CY// BX , BY// CX => YCXB # <BEC= < ACD.= w= < XAC = < AXE => DE=DX , so D lies on perp. bisector of EX And since D also lies on perpendicular bisector of AC, we conclude that ACXE is isosceles trapezoid (AC//EX, EA= CX) , so <<EAX = <CXE= < YBE = <BYE , which means YABE is cyclic