To construct a dodecahedron and comprehend it in a sphere, like the aforesaid and to prove that the side of the dodecahedron is the irrational straight line called apotome.

Line 2

Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out,

Line 3

let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, R, L, M, N, O respectively,

Line 4

let GK, HL, MH, NO be joined.

Line 5

let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively,

Line 6

and let RP, PS, TQ be their greater segments;

Line 7

from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube,
let them be made equal to RP, PS, TQ,

Line 8

and let UB, BW, WC, CV, VU be joined.

Line 9

I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular. For let RB, SB, VB be joined.

Line 10

Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment,
therefore the squares on PN, NR are triple of the square on RP.

Line 11

But PN is equal to NB, and PR to RU;

Line 12

therefore the squares on BN, NR are triple of the square on RU.

Line 13

But the square on BR is equal to the squares on BN, NR;
[I.47]

Line 14

therefore the square on BR is triple of the square on RU;
hence the squares on BR, RU are quadruple of the square on RU.

Line 15

But the square on BU is equal to the squares on BR, RU;

Line 16

therefore the square on BU is quadruple of the square on RU;

Line 17

therefore BU is double of RU

Line 18

But VU is also double of UR,
inasmuch as SR is also double of PR, that is, of RU;

Line 19

therefore BU is equal to UV.

Line 20

Similarly it can be proved that each of the straight lines BW, WC, CV is also equal to each of the straight lines BU, UV.

Line 21

Therefore the pentagon BUVCW is equilateral.

Line 22

I say next that it is also in one plane.
For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube,
and let XH, HW be joined;

Line 23

I say that XHW is a straight line.
For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment,
therefore, as HQ is to QT, so is QT to TH.

Line 24

But HQ is equal to HP,

Line 25

and QT to each of the straight lines TW, PX;

Line 26

therefore, as HP is to PX, so is WT to TH.

Line 27

And HP is parallel to TW, for each of them is at right angles to the plane BD;
[XI.6]

Line 28

and TH is parallel to PX, for each of them is at right angles to the plane BF.

Line 29

But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together
at one angle so that their corresponding sides are also parallel,
the remaining straight lines will be in a straight line;
[VI.32]
therefore XH is in a straight line with HW.

Line 30

But every straight line is in one plane;
[XI.1]
therefore the pentagon UBWCV is in one plane.

Line 31

I say next that it is also equiangular.
For, since the straight line NP has been cut in extreme and mean ratio at R,
and PR is the greater segment, while PR is equal to PS,
therefore NS has also been cut in extreme and mean ratio at P,

Line 32

and NP is the greater segment;
[XIII.5]

Line 33

therefore the squares on NS, SP are triple of the square on NP.
[XIII.4]

Line 34

But NP is equal to NB, and PS to SV;

Line 35

therefore the squares on NS, SV are triple of the square on NB;

Line 36

hence the squares on VS, SN, NB are quadruple of the square on NB.

Line 37

But the square on SB is equal to the squares on SN, NB;

Line 38

therefore the squares on BS, SV, that is, the square on
BV -- for the angle VSB is right --is quadruple of the square on NB;

Line 39

therefore VB is double of BN.

Line 40

But BC is also double of BN;

Line 41

therefore BV is equal to BC.

Line 42

And, since the two sides BU, UV are equal to the two sides BW, WC,
and the base BV is equal to the base BC,
therefore the angle BUV is equal to the angle BWC.
[I.8]

Line 43

Similarly we can prove that the angle UVC is also equal to the angle BWC;

Line 44

therefore the three angles BWC, BUV, UVC are equal to one another.

Line 45

But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular,
[XIII.7]
therefore the pentagon BUVCW is equiangular.

Line 46

And it was also proved equilateral;
therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube.

Line 47

Therefore, if we make the same construction in the case of each of the twelve sides of the cube,
a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons,
and which is called a dodecahedron.

Line 48

It is then required to comprehend it in the given sphere, and to prove that
one side of the dodecahedron is the irrational straight line called apotome.
For let XP be produced, and let the produced straight line be XZ; therefore PZ meets the diameter
of the cube, and they bisect one another, this has been proved in the last theorem but one of the
eleventh book.
[XI.38]

Line 49

Let them cut at Z;

Line 50

therefore Z is the center of the sphere which comprehends the cube,
and ZP is half of the side of the cube.

Line 51

`
Let UZ be joined.
`

`
Line 52
Now, since the straight line NS has been cut in extreme and mean ratio at P,
and NP is its greater segment, therefore the squares on NS, SP are triple of the square on NP.
[XIII.4]
`

`
Line 53
But NS is equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS.
`

`
Line 54
But further, PS is also equal to XU, since it is also equal to RP;
`

`
Line 55
therefore the squares on ZX, XU are triple of the square on NP.
`

`
Line 56
But the square on UZ is equal to the squares on ZX, XU;
`

`
Line 57
therefore the square on UZ is triple of the square on NP.
`

`
Line 58
But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube,
for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter
of the sphere is triple of the square on the side of the cube.
[XIII.15]
But, if whole is so related to whole, so is half to half also;
and NP is half of the side of the cube;
`

`
Line 59
therefore UZ is equal to the radius of the sphere which comprehends the cube.
`

`
Line 60
And Z is the center of the sphere which comprehends the cube;
`

`
Line 61
therefore the point U is on the surface of the sphere.
`

`
Line 62
Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere;
therefore the dodecahedron has been comprehended in the given sphere.
`

`
Line 63
I say next that the side of the dodecahedron is the irrational straight line called apotome.
For since, when NP has been cut in extreme and mean ratio, RP is the greater segment,
`

`
Line 64
and, when PO has been cut in extreme and mean ratio, PS is the greater segment,
`

`
Line 65
therefore, when the whole NO is cut in extreme and mean ratio, RS is the greater segment.
[Thus, since, as NP is to PR, so is PR to RN,
the same is true of the doubles also,
for parts have the same ratio as their equimultiples;
[V.15]
`

`
Line 66
therefore as NO is to RS, so is RS to the sum of NR, SO.
`

`
Line 67
But NO is greater than RS;
therefore RS is also greater than the sum of NR, SO;
`

`
Line 68
therefore NO has been cut in extreme and mean ratio,
and RS is its greater segment.]
`

`
Line 69
But RS is equal to UV;
`

`
Line 70
therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.
`

`
Line 71
And, since the diameter of the sphere is rational,
and the square on it is triple of the square on the side of the cube,
therefore NO, being a side of the cube, is rational.
`

`
Line 72
But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.
Therefore UV, being a side of the dodecahedron, is an irrational apotome.
[XIII.6]
`

`
Line 73
PORISM
From this it is manifest that, when the side of the cube is cut in extreme and mean ratio,
the greater segment is the side of the dodecahedron.
`

`
Q.E.D.`