Lecture
Theorem
The diagonal of the parallelepiped intersect at one point and the intersection point is divided in half.
Evidence
Consider any two diagonal parallelepiped, for example A1A3 `and A4A2`. Since the quadrangles A1A2A3A4 and A2A2`A3`A3 are parallelograms with the common side A2A3, their sides A1A4 and A2`A3` are parallel to each other, therefore, lie in the same plane. This plane intersects the planes of the opposite faces of the parallelepiped along parallel straight lines A1A2` and A4A3`. Therefore, the quadrilateral A4A1A2`A3` is a parallelogram. The diagonal of the parallelepiped A1A3` and A2A4` are the diagonals of this parallelogram. Therefore, they intersect and the intersection point O is divided in half.
Similarly, it is proved that the diagonals A1A3` and A2A4`, as well as the diagonals A1A3` and A3A1` intersect and the intersection point is divided in half. The theorem is proved.
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Stereometry
Terms: Stereometry