# Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected

Jump to navigation
Jump to search

## Theorem

In the words of Euclid:

*If a parallelepidedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane.*

(*The Elements*: Book $\text{XI}$: Proposition $28$)

## Proof

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 Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

- $\triangle CGF = \triangle CFB$

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 Book $\text{XI}$ Definition $10$: Similar Equal Solid Figures:

- 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.

$\blacksquare$

## Historical Note

This proof is Proposition $28$ of Book $\text{XI}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions