Solution:

Note that $Q$ is a parallelogram because each side is formed by joining the midpoints of two sides of a triangle, so it is parallel to and half the length of the base of the triangle. But the triangles corresponding to opposite sides have the same base. Hence opposite sides of $Q$ are parallel and equal. Similarly $Q'$.

Let the midpoints of the diagonals be $X$, $Y$. Take two adjacent side midpoints which are on the same side of the line $XY$. Suppose they are $M$, the midpoint of $AB$, and $N$, the midpoint of $BC$. Suppose also that $X$ is the midpoint of $BD$, and $Y$ the midpoint of $AC$. If $X$ does not lie on $AC$, then we may assume it lies on the same side of $AC$ as $M$ and $N$ (if not just consider the other two midpoints instead of $M$ and $N$). So the line parallel to $XY$ through $M$ cuts the altitude from $N$ of $NXY$. So $XYM$ has the same base $XY$ as $XYN$, but smaller height, so it has smaller area. Hence the two parallelograms also have different areas. Contradiction. So $X$ must lie on $AC$. But $AX$ bisects $ABD$ and $CX$ bisects $CBD$, so $AC$ bisects $ABD$ and $CBD$ and hence $ABCD$.