Solution:

Construct point $E$ on diagonal $BD$ so that $EX$ is parallel to $CD$. So $\angle BXE = \angle BCD = 90^\circ$. Also $\angle XBE = 45^\circ$ because $E$ is on diagonal $BD$. Therefore triangle $\triangle BEX$ is an isosceles right-angled triangle. Hence
$$EX = BX = YD.$$
Since segments $EX$ and $YD$ are parallel and equal in length, this implies that $EXDY$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, we deduce that the intersection of $XY$ and $DE$ is the midpoint of $XY$. Therefore the midpoint of $XY$ lies on line $DE$ (which is a segment of diagonal $BD$).

Alternative Solution A:
Let $ABCD$ be the unit square, with $A = (0,1)$, $B = (1,1)$, $C = (1,0)$ and $D = (0,0)$. Now let $a = BX = DY$. This means that $X = (1,1 - a)$ and $Y = (-a,0)$. Now let $Z$ be the midpoint of $XY$. We compute the coordinates of $Z$ to be
$$Z = \left(\frac{1 + (-a)}{2}, \frac{(1 - a) + 0}{2}\right).$$
The $x$ and $y$ coordinates of $Z$ are equal, therefore $Z$ lies on diagonal $BD$.

Alternative Solution B:
Let $s$ be the side-length of the square and let $a = BX = DY$. Let $Z$ be the midpoint of $XY$. Now we apply the converse of Menelaus' Theorem to traversal $BZD$ of $\triangle XYC$.
$$\frac{XZ}{ZY} \times \frac{YD}{DC} \times \frac{CB}{BX} = 1 \times \frac{a}{s} \times \frac{s}{a} = 1.$$
Therefore $Z$, $D$ and $B$ are colinear.