---
book: 13
number: 14
id: "XIII.14"
kind: "construction"
uses: ["[[book-11/proposition-12]]", "[[book-1/proposition-47]]", "[[book-1/proposition-4]]", "[[book-3/proposition-31]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.14"
license: "CC-BY-SA-4.0"
---

# XIII.14

*To construct an octahedron and comprehend it in a sphere*, *as in the preceding case*; *and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.*

## Proof

Let the diameter *AB* of the given sphere be set out, and let it be bisected at *C*; let the semicircle *ADB* be described on *AB*, let *CD* be drawn from *C* at right angles to *AB*, let *DB* be joined; let the square *EFGH*, having each of its sides equal to *DB*, be set out, let *HF*, *EG* be joined, from the point *K* let the straight line *KL* be set up at right angles to the plane of the square *EFGH* [[book-11/proposition-12|XI. 12]], and let it be carried through to the other side of the plane, as *KM*; from the straight lines *KL*, *KM* let *KL*, *KM* be respectively cut off equal to one of the straight lines *EK*, *FK*, *GK*, *HK*, and let *LE*, *LF*, *LG*, *LH*, *ME*, *MF*, *MG*, *MH* be joined.

Then, since *KE* is equal to *KH*, and the angle *EKH* is right, therefore the square on *HE* is double of the square on *EK*. [[book-1/proposition-47|I. 47]]

Again, since *LK* is equal to *KE*, and the angle *LKE* is right, therefore the square on *EL* is double of the square on *EK*. [*id.*]

But the square on *HE* was also proved double of the square on *EK*; therefore the square on *LE* is equal to the square on *EH*; therefore *LE* is equal to *EH*.

For the same reason *LH* is also equal to *HE*; therefore the triangle *LEH* is equilateral.

Similarly we can prove that each of the remaining triangles of which the sides of the square *EFGH* are the bases, and the points *L*, *M* the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles.

It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.

For, since the three straight lines *LK*, *KM*, *KE* are equal to one another, therefore the semicircle described on *LM* will also pass through *E*.

And for the same reason, if, *LM* remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points *F*, *G*, *H*, and the octahedron will have been comprehended in a sphere.

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

For, since *LK* is equal to *KM*, while *KE* is common, and they contain right angles, therefore the base *LE* is equal to the base *EM*. [[book-1/proposition-4|I. 4]]

And, since the angle *LEM* is right, for it is in a semicircle, [[book-3/proposition-31|III. 31]] therefore the square on *LM* is double of the square on *LE*. [[book-1/proposition-47|I. 47]]

Again, since *AC* is equal to *CB*, *AB* is double of *BC*.

But, as *AB* is to *BC*, so is the square on *AB* to the square on *BD*; therefore the square on *AB* is double of the square on *BD*.

But the square on *LM* was also proved double of the square on *LE*.

And the square on *DB* is equal to the square on *LE*, for *EH* was made equal to *DB*.

Therefore the square on *AB* is also equal to the square on *LM*; therefore *AB* is equal to *LM*.

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

Therefore the octahedron 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 double of the square on the side of the octahedron. Q. E. D.