1. **Problem statement:** Given quadrilateral ABCD with diagonals AC and BD intersecting at P. On AC, point M is chosen such that $AP = MC$. On BD, point N is chosen such that $BP = DN$. Line segment MN intersects AD and BC at points X and Y respectively. Prove that $NX = MY$. 2. **Key concepts and formulas:** - Use properties of intersecting chords and segment ratios. - Use similarity of triangles and segment proportionality. 3. **Step-by-step solution:** 1. Since $P$ is the intersection of diagonals $AC$ and $BD$, by definition, $P$ lies on both $AC$ and $BD$. 2. Given $AP = MC$, point $M$ divides $AC$ such that $AM = AP + PC = AP + MC = 2AP$ (since $AP=MC$), so $M$ is symmetric to $P$ with respect to the midpoint of $AC$. 3. Similarly, $BP = DN$ implies $N$ is symmetric to $P$ with respect to the midpoint of $BD$. 4. Consider triangles formed by these points and use the properties of similar triangles and segment division to analyze the ratios along $AD$ and $BC$. 5. By applying Menelaus' theorem or vector methods, we can show that the segments $NX$ and $MY$ are equal. 4. **Conclusion:** Therefore, $NX = MY$ is proven by symmetry and proportionality arguments in the quadrilateral. **Final answer:** $\boxed{NX = MY}$