1. Identify the given information and draw the diagram:
- $ABCD$ is a convex quadrilateral with $AB = CD$.
- $AC$ and $BD$ intersect at $O$.
- $X, Y, Z, T$ are midpoints of $BC, AD, AC, BD$ respectively.

2. **Prove that $XTYZ$ is a rhombus:**
- Since $X$ and $Y$ are midpoints of $BC$ and $AD$ respectively, $XY$ is parallel to $AB$ and $CD$ and $XY = \frac{1}{2}(AB + CD)$.
- Since $Z$ and $T$ are midpoints of $AC$ and $BD$ respectively, $ZT$ is parallel to $AC$ and $BD$ and $ZT = \frac{1}{2}(AC + BD)$.
- Given $AB = CD$, it follows that $XY = \frac{1}{2}(AB + CD) = AB = CD$.
- Similarly, $ZT = \frac{1}{2}(AC + BD)$.
- Since $X, Y, Z, T$ are midpoints, $XZ = YT$ and $XY = ZT$.
- Therefore, $XTYZ$ is a rhombus because all sides are equal and opposite sides are parallel.

3. **Prove that the circumcenter of $\triangle OZT$ lies on $XY$:**
- In a rhombus, the diagonals bisect each other at right angles.
- The diagonals of rhombus $XTYZ$ are $XZ$ and $YT$, and they intersect at the midpoint of both diagonals, which is the center of the rhombus.
- The circumcenter of $\triangle OZT$ is the point equidistant from $O, Z, T$.
- Since $O, Z, T$ are collinear with the center of the rhombus, the circumcenter of $\triangle OZT$ must lie on the perpendicular bisector of $ZT$.
- The perpendicular bisector of $ZT$ is the line $XY$ because $XY$ is the line that bisects the rhombus and is perpendicular to $ZT$.

Therefore, the circumcenter of $\triangle OZT$ lies on $XY$.

$\blacksquare$