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