Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected

Proof

 * Euclid-XI-28.png

Let the parallelepiped $AB$ be cut by the plane $CDEF$ through the diagonals $CF$ and $DE$ of opposite planes.

It is to be demonstrated that the parallelepiped $AB$ is bisected by the plane $CDEF$

From :
 * $\triangle CGF = \triangle CFB$

and:
 * $\triangle ADE = \triangle DEH$

while:
 * the parallelogram $CA$ is equal to the parallelogram $EB$

and:
 * the parallelogram $GE$ is equal to the parallelogram $CH$

as they are opposite.

Therefore from :
 * the prism contained by $\triangle CGF$ and $\triangle ADE$ and the three parallelograms $GE, AC, AE$

equals:
 * the prism contained by $\triangle CFB$ and $\triangle DEH$ and the three parallelograms $CH, BE, CE$.

Hence the result.