It's clear that $DC$ is one of the angular bisectors of $\angle{EDX}$, and by the fact that the orthocenter of a triangle is the isogonal conjugate of the circumcenter, with some angle chasing we get $EK$ is perpendicular to $DC$, so $DY$ is also an angular bisector of $\angle{EDX}$. therefore the problem is equivalent to proving $Y$ is the midpoint of arc $EDX$ in the circle $(EDX)$ let the radius of circle $(EDX)$ be $R$ and the radius of $(AEB)$ be $r$, $\angle{BDC}=\theta$ to prove this we first know that if $Y$ is the arc midpoint then \[DY=2R\sin({\frac{\angle{DEX}-\angle{DXE}}{2}}) =2R\sin{\angle{EXB}}=\frac{EX\sin{\angle{EXB}}}{\sin{2\theta}}=\frac{EB\sin{\angle{EBX}}}{\sin{2\theta}}=\frac{EB\cos{\theta}}{\sin{2\theta}}=\frac{EB}{2\sin{\theta}}\]while it's trivial that $r=\frac{EB}{2\sin{\theta}}$ and by parallelogram we know $Y$ is the midpoint of arc $EDX$ in the circle $(EDX)$ and we are done

Let $BX \cap CD = Z$. Let $\angle BDZ = \angle XDZ = \alpha$ Hence, $\angle BAC = \angle BDC = \alpha$ and $\angle BKE = 2 \cdot \angle BAE = 2 \alpha$ Note that since $K$ is the circumcenter of $\triangle ABE$, $KB=KE=KA \implies KB=KE=DY$ Also $\triangle BDZ \cong \triangle XDZ \implies BD=XD$ $\angle BKE + \angle KBD = \angle DEK = \angle EDY = 2\alpha + \angle YDX \implies \angle KBD = \angle YDX$ $\therefore \triangle YDX \cong \triangle KBD (SAS)$ $\angle YXD = \angle KDB = \angle KDE = \angle YED$ $\implies D, E, X, Y$ are concyclic.

We have $(DY, DX) \equiv (DY, DE) + (DE, DX) \equiv (EK, ED) + 2(DB, DC) \equiv (EK, ED) + (KB, KE) \equiv (KB, ED) \pmod \pi$. But $DY = KE = KB$ and $DB = DX$ then $\triangle DXY \cong \triangle BDK$. So $(XD, XY) \equiv (DB, DK) \equiv (ED, EY) \pmod \pi$ or $X, Y, D, E$ lie on a circle