$ABCD$ is a quadrilateral such that $AB$ is not parallel to $CD$, and $BC$ is not parallel to $AD$. Variable points $P, Q, R, S$ are taken on $AB, BC, CD, DA$ respectively so that $PQRS$ is a parallelogram. Find the locus of its center.
1. Consider the quadrilateral \(ABCD\) where \(AB\) is not parallel to \(CD\) and \(BC\) is not parallel to \(AD\). We need to find the locus of the center of a parallelogram \(PQRS\) formed by points \(P, Q, R, S\) on \(AB, BC, CD, DA\) respectively.

2. Begin by selecting an arbitrary point \(P\) on \(AB\). Draw lines through \(P\) parallel to \(AC\) and \(BD\). Let these lines intersect \(BC\) and \(AD\) at points \(Q\) and \(S\) respectively.

3. To complete the parallelogram \(PQRS\), find point \(R\) on \(CD\) such that \(QR \parallel BD \parallel PS\). This ensures that \(RS \parallel AC \parallel PQ\).

4. We need to show that the midpoint \(M\) of \(QS\) (which is the center of parallelogram \(PQRS\)) lies on the segment \(XY\), where \(X\) and \(Y\) are the midpoints of \(AC\) and \(BD\) respectively.

5. By the properties of parallelograms and the given construction, we have:
   \[
   \frac{AS}{SD} = \frac{AP}{PB} = \frac{CQ}{QB} = \frac{CR}{RD}
   \]
   This implies that the ratios of the segments are equal.

6. Using the ERIQ theorem (Equal Ratios In Quadrilateral), we know that:
   \[
   \frac{BQ}{QC} = \frac{DS}{SA} \quad \text{and} \quad \frac{BY}{YD} = \frac{QM}{MS} = \frac{CX}{XA}
   \]
   This theorem helps us establish that the midpoint \(M\) of \(QS\) lies on the line segment \(XY\).

7. Therefore, the locus of the center of the parallelogram \(PQRS\) is the line segment \(XY\).

The final answer is the line segment \( \boxed{ XY } \).