---
book: 11
number: 38
id: "XI.38"
kind: "theorem"
uses: ["[[book-1/proposition-29]]", "[[book-1/proposition-4]]", "[[book-1/proposition-14]]", "[[book-11/proposition-9]]", "[[book-1/proposition-33]]", "[[book-1/proposition-15]]", "[[book-1/proposition-26]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.38"
license: "CC-BY-SA-4.0"
---

# XI.38

*If the sides of the opposite planes of a cube be bisected*, *and planes be carried through the points of section*, *the common section of the planes and the diameter of the cube bisect one another.*

## Proof

For let the sides of the opposite planes *CF*, *AH* of the cube *AF* be bisected at the points *K*, *L*, *M*, *N*, *O*, *Q*, *P*, *R*, and through the points of section let the planes *KN*, *OR* be carried; let *US* be the common section of the planes, and *DG* the diameter of the cube *AF*.

I say that *UT* is equal to *TS*, and *DT* to *TG*.

For let *DU*, *UE*, *BS*, *SG* be joined.

Then, since *DO* is parallel to *PE*, the alternate angles *DOU*, *UPE* are equal to one another. [[book-1/proposition-29|I. 29]]

And, since *DO* is equal to *PE*, and *OU* to *UP*, and they contain equal angles, therefore the base *DU* is equal to the base *UE*, the triangle *DOU* is equal to the triangle *PUE*, and the remaining angles are equal to the remaining angles; [[book-1/proposition-4|I. 4]] therefore the angle *OUD* is equal to the angle *PUE*.

For this reason *DUE* is a straight line. [[book-1/proposition-14|I. 14]]

For the same reason, *BSG* is also a straight line, and *BS* is equal to *SG*.

Now, since *CA* is equal and parallel to *DB*, while *CA* is also equal and parallel to *EG*, therefore *DB* is also equal and parallel to *EG*. [[book-11/proposition-9|XI. 9]]

And the straight lines *DE*, *BG* join their extremities; therefore *DE* is parallel to *BG*. [[book-1/proposition-33|I. 33]]

Therefore the angle *EDT* is equal to the angle *BGT*, for they are alternate; [[book-1/proposition-29|I. 29]] and the angle *DTU* is equal to the angle *GTS*. [[book-1/proposition-15|I. 15]]

Therefore *DTU*, *GTS* are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, *DU* equal to *GS*, for they are the halves of *DE*, *BG*; therefore they will also have the remaining sides equal to the remaining sides. [[book-1/proposition-26|I. 26]]

Therefore *DT* is equal to *TG*, and *UT* to *TS*.

Therefore etc. Q. E. D.