---
book: 11
number: 28
id: "XI.28"
kind: "theorem"
uses: ["[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.28"
license: "CC-BY-SA-4.0"
---

# XI.28

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

## Proof

For let the parallelepipedal solid *AB* be cut by the plane *CDEF* through the diagonals *CF*, *DE* of opposite planes; I say that the solid *AB* will be bisected by the plane *CDEF*.

For, since the triangle *CGF* is equal to the triangle *CFB*, [[book-1/proposition-34|I. 34]] and *ADE* to *DEH*, while the parallelogram *CA* is also equal to the parallelogram *EB*, for they are opposite, and *GE* to *CH*, therefore the prism contained by the two triangles *CGF*, *ADE* and the three parallelograms *GE*, *AC*, *CE* is also equal to the prism contained by the two triangles *CFB*, *DEH* and the three parallelograms *CH*, *BE*, *CE*; for they are contained by planes equal both in multitude and in magnitude. [[book-11/definitions#Definition 10|XI. Def. 10]]

Hence the whole solid *AB* is bisected by the plane *CDEF*. Q. E. D.