I have a box $ABCDA'B'C'D'$ where its lower base is $ABCD$ and its higher base is $A'B'C'D'$. Let $O$ be intersection of the diagonals of $ABCD$ and $O'$ be the intersection of the diagonals of $A'B'C'D'$.
How can I prove that $O'O || A'A$? I know it's easy to see it but I don't have a clue in how to prove it geometrically.
The quadrilateral $ACC'A'$ is a rectangle.
Indeed, by the definition of a cube $AA'$ is orthogonal to both $AB$ and $AD$. Therefore, since if a line is orthogonal to two transverse lines lying in a plane, then it is orthogonal to the whole plane, $AA'$ is orthogonal to the face $ABCD$. Analogous arguments show that $CC'$ is orthogonal to the same face $ABCD$. Therefore $AA'$ and $CC'$ are parallel to each other and so they lie in a common plane $AA'CC'$. Combined with the fact that $AA'=CC'$ (as edges of a cube) we conclude that $ACC'A'$ is a parallelogram. However, since $AA'$ is orthogonal to face $ABCD$, it is orthogonal to any line lying in $ABCD$. In particular, $AA'$ is orthogonal to $AC$. Thus, $ACC'A'$ is a parallelogram with a right angle, so it is a rectangle.
Consequently $AC = A'C'$ and $AC \,\|\, A'C'$. The points $O$ and $O'$ are the midpoints of the segments $AC$ and $AC'$ respectively. Therefore $$AO = \frac{1}{2} AC = \frac{1}{2} A'C' = A'O$$ and $AO \,\|\, A'O'$. Therefore $AOO'A'$ is a parallelogram (in fact a rectangle) and consequently $AA' = OO'$ and $AA' \,\|\, OO'$. Moreover, $OO'$ is orthogonal to both faces $ABCD$ and $A'B'C'D'$.