XXVII Olimpiada Matemática Rioplatense

Let $N$ be the intersection point of the diagonals of $ABCD$. As $ABCD$ is a parallelogram, we have that $N$ is the midpoint of $AC$ and the midpoint of $BD$.



Since $P$ is the center of the square $BDXY$ and $N$ is the midpoint of its side $BD$, then $DN=NP$ (both equal to a half of $BD$) and $D\hat{N}P=90^{\circ }$. Similarly, $AN=NQ$ and $A\hat{N}Q=90^{\circ }$. Then, $A\hat{N}P=90^{\circ }+A\hat{N}D=Q\hat{N}D$. We conclude that the triangles $ANP$ and $QND$ are congruent (by side-angle-side criterion) and, therefore, $AP=DQ$.