---
book: 13
number: 15
id: "XIII.15"
kind: "construction"
uses: ["[[book-1/proposition-47]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.15"
license: "CC-BY-SA-4.0"
---

# XIII.15

*To construct a cube and comprehend it in a sphere*, *like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.*

## Proof

Let the diameter *AB* of the given sphere be set out, and let it be cut at *C* so that *AC* is double of *CB*; let the semicircle *ADB* be described on *AB*, let *CD* be drawn from *C* at right angles to *AB*, and let *DB* be joined; let the square *EFGH* having its side equal to *DB* be set out, from *E*, *F*, *G*, *H* let *EK*, *FL*, *GM*, *HN* be drawn at right angles to the plane of the square *EFGH*, from *EK*, *FL*, *GM*, *HN* let *EK*, *FL*, *GM*, *HN* respectively be cut off equal to one of the straight lines *EF*, *FG*, *GH*, *HE*, and let *KL*, *LM*, *MN*, *NK* be joined; therefore the cube *FN* has been constructed which is contained by six equal squares.

It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.

For let *KG*, *EG* be joined.

Then, since the angle *KEG* is right, because *KE* is also at right angles to the plane *EG* and of course to the straight line *EG* also, [[book-11/definitions#Definition 3|XI. Def. 3]] therefore the semicircle described on *KG* will also pass through the point *E*.

Again, since *GF* is at right angles to each of the straight lines *FL*, *FE*, *GF* is also at right angles to the plane *FK*; hence also, if we join *FK*, *GF* will be at right angles to *FK*; and for this reason again the semicircle described on *GK* will also pass through *F*.

Similarly it will also pass through the remaining angular points of the cube.

If then, *KG* remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere.

I say next that it is also comprehended in the given sphere.

For, since *GF* is equal to *FE*, and the angle at *F* is right, therefore the square on *EG* is double of the square on *EF*.

But *EF* is equal to *EK*; therefore the square on *EG* is double of the square on *EK*; hence the squares on *GE*, *EK*, that is the square on *GK* [[book-1/proposition-47|I. 47]], is triple of the square on *EK*.

And, since *AB* is triple of *BC*, while, as *AB* is to *BC*, so is the square on *AB* to the square on *BD*, therefore the square on *AB* is triple of the square on *BD*.

But the square on *GK* was also proved triple of the square on *KE*.

And *KE* was made equal to *DB*; therefore *KG* is also equal to *AB*.

And *AB* is the diameter of the given sphere; therefore *KG* is also equal to the diameter of the given sphere.

Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube. Q. E. D.