---
book: 13
number: 13
id: "XIII.13"
kind: "construction"
uses: ["[[book-4/proposition-2]]", "[[book-3/proposition-1]]", "[[book-11/proposition-12]]", "[[book-1/proposition-4]]", "[[book-13/proposition-12]]", "[[book-6/proposition-8]]", "[[book-6/proposition-17]]", "[[book-3/proposition-31]]", "[[book-6/proposition-4]]", "[[book-6/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.13"
license: "CC-BY-SA-4.0"
---

# XIII.13

*To construct a pyramid*, *to comprehend it in a given sphere*, *and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.*

## Proof

Let the diameter *AB* of the given sphere be set out, and let it be cut at the point *C* so that *AC* is double of *CB*; let the semicircle *ADB* be described on *AB*, let *CD* be drawn from the point *C* at right angles to *AB*, and let *DA* be joined; let the circle *EFG* which has its radius equal to *DC* be set out, let the equilateral triangle *EFG* be inscribed in the circle *EFG*, [[book-4/proposition-2|IV. 2]] let the centre of the circle, the point *H*, be taken, [[book-3/proposition-1|III. 1]] let *EH*, *HF*, *HG* be joined; from the point *H* let *HK* be set up at right angles to the plane of the circle *EFG*, [[book-11/proposition-12|XI. 12]] let *HK* equal to the straight line *AC* be cut off from *HK*, and let *KE*, *KF*, *KG* be joined.

Now, since *KH* is at right angles to the plane of the circle *EFG*, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle *EFG*. [[book-11/definitions#Definition 3|XI. Def. 3]]

But each of the straight lines *HE*, *HF*, *HG* meets it: therefore *HK* is at right angles to each of the straight lines *HE*, *HF*, *HG*.

And, since *AC* is equal to *HK*, and *CD* to *HE*, and they contain right angles, therefore the base *DA* is equal to the base *KE*. [[book-1/proposition-4|I. 4]]

For the same reason each of the straight lines *KF*, *KG* is also equal to *DA*; therefore the three straight lines *KE*, *KF*, *KG* are equal to one another.

And, since *AC* is double of *CB*, therefore *AB* is triple of *BC*.

But, as *AB* is to *BC*, so is the square on *AD* to the square on *DC*, as will be proved afterwards.

Therefore the square on *AD* is triple of the square on *DC*.

But the square on *FE* is also triple of the square on *EH*, [[book-13/proposition-12|XIII. 12]] and *DC* is equal to *EH*; therefore *DA* is also equal to *EF*.

But *DA* was proved equal to each of the straight lines *KE*, *KF*, *KG*; therefore each of the straight lines *EF*, *FG*, *GE* is also equal to each of the straight lines *KE*, *KF*, *KG*; therefore the four triangles *EFG*, *KEF*, *KFG*, *KEG* are equilateral.

Therefore a pyramid has been constructed out of four equilateral triangles, the triangle *EFG* being its base and the point *K* its vertex.

It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

For let the straight line *HL* be produced in a straight line with *KH*, and let *HL* be made equal to *CB*.

Now, since, as *AC* is to *CD*, so is *CD* to *CB*, [[book-6/proposition-8|VI. 8, Por.]] while *AC* is equal to *KH*, *CD* to *HE*, and *CB* to *HL*, therefore, as *KH* is to *HE*, so is *EH* to *HL*; therefore the rectangle *KH*, *HL* is equal to the square on *EH*. [[book-6/proposition-17|VI. 17]]

And each of the angles *KHE*. *EHL* is right; therefore the semicircle described on *KL* will pass through *E* also. cf. [[book-6/proposition-8|VI. 8]], [[book-3/proposition-31|III. 31]].

If then, *KL* 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*, since, if *FL*, *LG* be joined, the angles at *F*, *G* similarly become right angles; and the pyramid will be comprehended in the given sphere.

For *KL*, the diameter of the sphere, is equal to the diameter *AB* of the given sphere, inasmuch as *KH* was made equal to *AC*, and *HL* to *CB*.

I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid

For, since *AC* is double of *CB*, therefore *AB* is triple of *BC*; and, convertendo, *BA* is one and a half times *AC*.

But, as *BA* is to *AC*, so is the square on *BA* to the square on *AD*.

Therefore the square on *BA* is also one and a half times the square on *AD*.

And *BA* is the diameter of the given sphere, and *AD* is equal to the side of the pyramid.

Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.

Lemma. It is to be proved that, as *AB* is to *BC*, so is the square on *AD* to the square on *DC*.

For let the figure of the semicircle be set out, let *DB* be joined, let the square *EC* be described on *AC*, and let the parallelogram *FB* be completed.

Since then, because the triangle *DAB* is equiangular with the triangle *DAC*, as *BA* is to *AD*, so is *DA* to *AC*, [[book-6/proposition-8|VI. 8]], [[book-6/proposition-4|VI. 4]] therefore the rectangle *BA*, *AC* is equal to the square on *AD*. [[book-6/proposition-17|VI. 17]]

And since, as *AB* is to *BC*, so is *EB* to *BF*, [[book-6/proposition-1|VI. 1]] and *EB* is the rectangle *BA*, *AC*, for *EA* is equal to *AC*, and *BF* is the rectangle *AC*, *CB*, therefore, as *AB* is to *BC*, so is the rectangle *BA*, *AC* to the rectangle *AC*, *CB*.

And the rectangle *BA*, *AC* is equal to the square on *AD*, and the rectangle *AC*, *CB* to the square on *DC*, for the perpendicular *DC* is a mean proportional between the segments *AC*, *CB* of the base, because the angle *ADB* is right. [[book-6/proposition-8|VI. 8, Por.]]

Therefore, as *AB* is to *BC*, so is the square on *AD* to the square on *DC*. Q. E. D.