Solution:

First solution. Let $M$ be the midpoint of the segment $AB$. Let the parallel to $DY$ through $M$ meet $BD$ at $G$ and the parallel to $BD$ through $A$ at $H$. From $\overline{AM} = \overline{MB}$ one easily concludes $\overline{HM} = \overline{MG}$, and since $DY \perp DB$ is assumed, the quadrilateral $DGHY$ is a rectangle. Both facts together show $\overline{DM} = \overline{MY}$, and in the same way one shows $\overline{DM} = \overline{MX}$. Since the trapezoid $ABCD$ is isosceles, we further have $\overline{DM} = \overline{MC}$, and combining the last three equations immediately shows that the four points $C, D, X, Y$ lie on a common circle about $M$.

Second solution. Complete the triangle $DAB$ to a parallelogram $DAZB$. By the construction of $X$ and $Y$, these two points lie on $ZB$ and $ZA$ respectively, and we have $\angle DXZ = \angle ZYD = 90^{\circ}$. Thales' theorem now yields that $X$ and $Y$ lie on the circle with diameter $DZ$. The center of this circle is at the same time the midpoint of the segment $DZ$, and since in parallelograms the diagonals bisect each other, this coincides with the midpoint of the segment $AB$. Since the trapezoid $ABCD$ is now assumed to be isosceles, we consequently have $\overline{DM} = \overline{MC}$, which shows that the point $C$ also lies on the circle considered above.

Third solution (sketched, based on an idea of Andreas Gross). Let $F = (ABD)$ be the area of the triangle $ABD$. We draw about $D$ a circle $\omega$ with radius $\sqrt{2F}$ and invert with respect to it. Those familiar with the technique can easily convince themselves that $\overline{DX'} = \overline{DA}$ as well as $\overline{DY'} = \overline{DB}$ hold, and that from this $(DX'Y') = F$ follows. Let now $C^*$ be the intersection point of the line $DC$ and $X'Y'$, and let $H$ be the foot of the perpendicular dropped from $D$ onto $AB$. Now the points $X', Y'$ have distances $\overline{AH}, \overline{BH}$ from the line $CD$, and from this we obtain $(DX'Y') = (DC^*Y') - (DC^*X') = \frac{1}{2} \overline{DC^*} \cdot (\overline{HB} - \overline{AH}) = \frac{1}{2} \overline{DC^*} \cdot \overline{DC}$. In connection with an equation already given, it follows from this that $\overline{DC^*} \cdot \overline{DC} = 2F$, from which, with the help of a small positional consideration, one obtains $C^* = C'$. Accordingly, the three points $C', X', Y'$ lie on a common line, and one easily convinces oneself that this line cannot pass through $D$. This implies the claim.