Let $(AYD)\cap CD=Z\neq D$. Since $DZ\parallel AY$, $AD=YZ$. Extend $\overrightarrow{DC}$ to $E$ so that $CE=XY$. Note that $$DE=DC+CE=DC+(AB-CD)=AB.$$Since $DE\parallel AB$, $DEBA$ is a parallelogram. Thus, $AD=YZ=BE$. Note that $\angle ZYA=\angle DAY=\angle DEB$ since $ADZY$ is isosceles trapezoid. Now, we know that $XY=CE, YZ=EB, \angle XYZ=\angle CEB$, therefore $$\triangle XYZ\cong\triangle CEB\implies \angle YXZ=\angle ECB.$$Thus, $Z,C,B,X$ concyclic. Hence, $(AYD)\cap(BXC)=Z$ which lies on $CD$.

Shorter solution. Let $E$ be point on $AB$ such that $AE=DC$ and let $(AYD)\cap DC=T$. So we need to prove $T\in (BXC)$ or $TX=BC$. Now from cosine law we get $BC^2=EC^2+EB^2-2EC\cdot EB\cdot \cos\angle BEC$ and $TX^2=TY^2+XY^2-2TY\cdot XY\cdot \cos \angle TYX$. And since $EC=AD=TY, EB=AB-AE=AB-DC=XY$ and $\angle BEC=\angle BAD=\angle BEC$, we get $BC^2=TX^2$ and the result follows.

Let $Z$ be a point on $CD$ such that $DZ=AX$. Obviously $CZ=BY$. Define $T$ to be the point on $CD$ such that $XYTZ$ is an isosceles trapezoid. Easily see that $DAYT$ and $XBCT$ are cyclic.