---
book: 13
number: 17
id: "XIII.17"
kind: "construction"
uses: ["[[book-13/proposition-4]]", "[[book-1/proposition-47]]", "[[book-11/proposition-6]]", "[[book-6/proposition-32]]", "[[book-11/proposition-1]]", "[[book-13/proposition-5]]", "[[book-1/proposition-8]]", "[[book-13/proposition-7]]", "[[book-11/proposition-38]]", "[[book-13/proposition-15]]", "[[book-5/proposition-15]]", "[[book-13/proposition-6]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.17"
license: "CC-BY-SA-4.0"
---

# XIII.17

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

## Proof

Let *ABCD*, *CBEF*, two planes of the aforesaid cube at right angles to one another, be set out, let the sides *AB*, *BC*, *CD*, *DA*, *EF*, *EB*, *FC* be bisected at *G*, *H*, *K*, *L*, *M*, *N*, *O* respectively, let *GK*, *HL*, *MH*, *NO* be joined, let the straight lines *NP*, *PO*, *HQ* be cut in extreme and mean ratio at the points *R*, *S*, *T* respectively, and let *RP*, *PS*, *TQ* be their greater segments; 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*, and let *UB*, *BW*, *WC*, *CV*, *VU* be joined.

I say that the pentagon *UBWCV* is equilateral, and in one plane, and is further equiangular.

For let *RB*, *SB*, *VB* be joined.

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*. [[book-13/proposition-4|XIII. 4]]

But *PN* is equal to *NB*, and *PR* to *RU*; therefore the squares on *BN*, *NR* are triple of the square on *RU*.

But the square on *BR* is equal to the squares on *BN*, *NR*; [[book-1/proposition-47|I. 47]] 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*.

But the square on *BU* is equal to the squares on *BR*, *RU*; therefore the square on *BU* is quadruple of the square on *RU*; therefore *BU* is double of *RU*.

But *VU* is also double of *UR*, inasmuch as *SR* is also double of *PR*, that is, of *RU*; therefore *BU* is equal to *UV*.

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*.

Therefore the pentagon *BUVCW* is equilateral.

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; 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*.

But *HQ* is equal to *HP*, and *QT* to each of the straight lines *TW*, *PX*; therefore, as *HP* is to *PX*, so is *WT* to *TH*.

And *HP* is parallel to *TW*, for each of them is at right angles to the plane *BD*; [[book-11/proposition-6|XI. 6]] and *TH* is parallel to *PX*, for each of them is at right angles to the plane *BF*. [*id*.]

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; [[book-6/proposition-32|VI. 32]] therefore *XH* is in a straight line with *HW*.

But every straight line is in one plane; [[book-11/proposition-1|XI. 1]] therefore the pentagon *UBWCV* is in one plane.

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*, and *NP* is the greater segment; [[book-13/proposition-5|XIII. 5]] therefore the squares on *NS*, *SP* are triple of the square on *NP*. [[book-13/proposition-4|XIII. 4]]

But *NP* is equal to *NB*, and *PS* to *SV*; therefore the squares on *NS*, *SV* are triple of the square on *NB*; hence the squares on *VS*, *SN*, *NB* are quadruple of the square on *NB*.

But the square on *SB* is equal to the squares on *SN*, *NB*; 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*; therefore *VB* is double of *BN*.

But *BC* is also double of *BN*; therefore *BV* is equal to *BC*.

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*. [[book-1/proposition-8|I. 8]]

Similarly we can prove that the angle *UVC* is also equal to the angle *BWC*; therefore the three angles *BWC*, *BUV*, *UVC* are equal to one another.

But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [[book-13/proposition-7|XIII. 7]] therefore the pentagon *BUVCW* is equiangular.

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

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.

It is then required to comprehend it in the given sphere, and to prove that the 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, for this has been proved in the last theorem but one of the eleventh book. [[book-11/proposition-38|XI. 38]]

Let them cut at *Z*; therefore *Z* is the centre of the sphere which comprehends the cube, and *ZP* is half of the side of the cube.

Let *UZ* be joined.

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*. [[book-13/proposition-4|XIII. 4]]

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

But further *PS* is also equal to *XU*, since it is also equal to *RP*; therefore the squares on *ZX*, *XU* are triple of the square on *NP*.

But the square on *UZ* is equal to the squares on *ZX*, *XU*; therefore the square on *UZ* is triple of the square on *NP*.

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. [[book-13/proposition-15|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; therefore *UZ* is equal to the radius of the sphere which comprehends the cube.

And *Z* is the centre of the sphere which comprehends the cube; therefore the point *U* is on the surface of the sphere.

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.

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, and, when *PO* has been cut in extreme and mean ratio, *PS* is the greater segment, 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; [[book-5/proposition-15|V. 15]] therefore as *NO* is to *RS*, so is *RS* to the sum of *NR*, *SO*.

But *NO* is greater than *RS*; therefore *RS* is also greater than the sum of *NR*, *SO*; therefore *NO* has been cut in extreme and mean ratio, and *RS* is its greater segment.]

But *RS* is equal to *UV*; therefore, when *NO* is cut in extreme and mean ratio, *UV* is the greater segment.

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.

[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. [[book-13/proposition-6|XIII. 6]]

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.