67th Romanian Mathematical Olympiad

Notice that the points $H$ and $K$ are on the diagonal $AC$, because $AC$ is perpendicular on $BE$ and $DE$. Also, $H$ and $K$ are on the altitudes from $E$ in the two triangles, which are perpendicular on the sides of the initial square.

It follows that the triangle $EHK$ is right and isosceles, so $H$ and $K$ are symmetric with respect of the square's center. Since $B$, $D$ are also symmetric with respect of the square's center, it follows that $DKBH$ is a parallelogram, whence the conclusion.