Proposition 13

Line 1

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.

Line 2

Let the diameter AB of the given sphere be set out.

Line 3

and let it be cut at the point C so that AC is double of CB;

Line 4

let the semicircle ADB be described on AB,

Line 5

let CD be drawn from the point C at right angles to AB,

Line 6

and let DA be joined;

Line 7

let the circle EFG which has its radius equal to DC be set out,

Line 8

let the equilateral triangle EFG be inscribed in the circle EFG, [IV.2]

Line 9

let the center of the circle, the point H, be taken, [III.1]

Line 10

let EH, HF, HG be joined;

Line 11

from the point H let HK be set up at right angles to the plane of the circle EFG, [XI.12] let HK equal to the straight line AC be cut off from HK,

Line 12

and let KE, KF, KG be joined.

Line 13

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. [XI. def.3]

Line 14

But each of the straight lines HE, HF, HG meets it:

Line 15

therefore HK is at right angles to each of the straight lines HE, HF, HG.

Line 16

And, since AC is equal to HK, and CD to HE, and they contain right angles,

Line 17

therefore the base DA is equal to the base KE.

Line 18

For the same reason each of the straight lines KF, KG is also equal to DA;

Line 19

therefore the three straight lines KE, KF, KG are equal to one another.

Line 20

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

Line 21

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.

Line 22

But the square on FE is also triple of the square on EH, [XIII.12] and DC is equal to EH: therefore DA is also equal to EF.

Line 23

But DA was proved equal to each of the straitght lines KE, KF, KG;

Line 24

therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG;

Line 25

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.

Line 26

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.

Line 27

Now, since as AC is to CD, so CD to CB, while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; [VI.8]

Line 28

therefore the rectangle KH, HL is equal to square on EH. [VI.17]

Line 29

And each of angles KHE, EHL is right;

Line 30

therefore the semicircle described on KL will pass through E also. [VI.8]

Line 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.

Line 32

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

Line 33

I say next that 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.

Line 34

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.