---
book: 13
number: 18
id: "XIII.18"
kind: "construction"
uses: ["[[book-6/proposition-8]]", "[[book-13/proposition-13]]", "[[book-13/proposition-15]]", "[[book-13/proposition-14]]", "[[book-13/proposition-16]]", "[[book-4/proposition-15]]", "[[book-13/proposition-10]]", "[[book-13/proposition-17]]", "[[book-6/proposition-4]]", "[[book-6/proposition-20]]", "[[book-13/proposition-9]]", "[[book-11/proposition-21]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.18"
license: "CC-BY-SA-4.0"
---

# XIII.18

*To set out the sides of the five figures and to compare them with one another.*

## Proof

Let *AB*, the diameter of the given sphere, be set out, and let it be cut at *C* so that *AC* is equal to *CB*, and at *D* so that *AD* is double of *DB*; let the semicircle *AEB* be described on *AB*, from *C*, *D* let *CE*, *DF* be drawn at right angles to *AB*, and let *AF*, *FB*, *EB* be joined.

Then, since *AD* is double of *DB*, therefore *AB* is triple of *BD*.

Convertendo, therefore, *BA* is one and a half times *AD*.

But, as *BA* is to *AD*, so is the square on *BA* to the square on *AF*, [[book-5/definitions#Definition 9|V. Def. 9]], [[book-6/proposition-8|VI. 8]] for the triangle *AFB* is equiangular with the triangle *AFD*; therefore the square on *BA* is one and a half times the square on *AF*.

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [[book-13/proposition-13|XIII. 13]]

And *AB* is the diameter of the sphere; therefore *AF* is equal to the side of the pyramid.

Again, since *AD* is double of *DB*, therefore *AB* is triple of *BD*.

But, as *AB* is to *BD*, so is the square on *AB* to the square on *BF*; [[book-6/proposition-8|VI. 8]], [[book-5/definitions#Definition 9|V. Def. 9]] therefore the square on *AB* is triple of the square on *BF*.

But the square on the diameter of the sphere is also triple of the square on the side of the cube. [[book-13/proposition-15|XIII. 15]]

And *AB* is the diameter of the sphere; therefore *BF* is the side of the cube.

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

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

But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [[book-13/proposition-14|XIII. 14]]

And *AB* is the diameter of the given sphere; therefore *BE* is the side of the octahedron.

Next, let *AG* be drawn from the point *A* at right angles to the straight line *AB*, let *AG* be made equal to *AB*, let *GC* be joined, and from *H* let *HK* be drawn perpendicular to *AB*.

Then, since *GA* is double of *AC*, for *GA* is equal to *AB*, and, as *GA* is to *AC*, so is *HK* to *KC*, therefore *HK* is also double of *KC*.

Therefore the square on *HK* is quadruple of the square on *KC*; therefore the squares on *HK*, *KC*, that is, the square on *HC*, is five times the square on *KC*.

But *HC* is equal to *CB*; therefore the square on *BC* is five times the square on *CK*.

And, since *AB* is double of *CB*, and, in them, *AD* is double of *DB*, therefore the remainder *BD* is double of the remainder *DC*.

Therefore *BC* is triple of *CD*; therefore the square on *BC* is nine times the square on *CD*.

But the square on *BC* is five times the square on *CK*; therefore the square on *CK* is greater than the square on *CD*; therefore *CK* is greater than *CD*.

Let *CL* be made equal to *CK*, from *L* let *LM* be drawn at right angles to *AB*, and let *MB* be joined.

Now, since the square on *BC* is five times the square on *CK*, and *AB* is double of *BC*, and *KL* double of *CK*, therefore the square on *AB* is five times the square on *KL*.

But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [[book-13/proposition-16|XIII. 16, Por.]]

And *AB* is the diameter of the sphere; therefore *KL* is the radius of the circle from which the icosahedron has been described; therefore *KL* is a side of the hexagon in the said circle. [[book-4/proposition-15|IV. 15, Por.]]

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [[book-13/proposition-16|XIII. 16, Por.]] and *AB* is the diameter of the sphere, while *KL* is a side of the hexagon, and *AK* is equal to *LB*, therefore each of the straight lines *AK*, *LB* is a side of the decagon inscribed in the circle from which the icosahedron has been described.

And, since *LB* belongs to a decagon, and *ML* to a hexagon, for *ML* is equal to *KL*, since it is also equal to *HK*, being the same distance from the centre, and each of the straight lines *HK*, *KL* is double of *KC*, therefore *MB* belongs to a pentagon. [[book-13/proposition-10|XIII. 10]]

But the side of the pentagon is the side of the icosahedron; [[book-13/proposition-16|XIII. 16]] therefore *MB* belongs to the icosahedron.

Now, since *FB* is a side of the cube, let it be cut in extreme and mean ratio at *N*, and let *NB* be the greater segment; therefore *NB* is a side of the dodecahedron. [[book-13/proposition-17|XIII. 17, Por.]]

And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side *AF* of the pyramid, double of the square on the side *BE* of the octahedron and triple of the side *FB* of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.

Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube; and the square on the side of the octahedron is one and a half times the square on the side of the cube.

The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.

But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [[book-13/proposition-16|XIII. 16]] and the other an apotome [[book-13/proposition-17|XIII. 17]].

That the side *MB* of the icosahedron is greater than the side *NB* of the dodecahedron we can prove thus.

For, since the triangle *FDB* is equiangular with the triangle *FAB*, [[book-6/proposition-8|VI. 8]] proportionally, as *DB* is to *BF*, so is *BF* to *BA*. [[book-6/proposition-4|VI. 4]]

And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [[book-5/definitions#Definition 9|V. Def. 9]], [[book-6/proposition-20|VI. 20, Por.]] therefore, as *DB* is to *BA*, so is the square on *DB* to the square on *BF*; therefore, inversely, as *AB* is to *BD*, so is the square on *FB* to the square on *BD*.

But *AB* is triple of *BD*; therefore the square on *FB* is triple of the square on *BD*.

But the square on *AD* is also quadruple of the square on *DB*, for *AD* is double of *DB*; therefore the square on *AD* is greater than the square on *FB*; therefore *AD* is greater than *FB*; therefore *AL* is by far greater than *FB*.

And, when *AL* is cut in extreme and mean ratio, *KL* is the greater segment, inasmuch as *LK* belongs to a hexagon, and *KA* to a decagon; [[book-13/proposition-9|XIII. 9]] and, when *FB* is cut in extreme and mean ratio, *NB* is the greater segment; therefore *KL* is greater than *NB*.

But *KL* is equal to *LM*; therefore *LM* is greater than *NB*.

Therefore *MB*, which is a side of the icosahedron, is by far greater than *NB* which is a side of the dodecahedron. Q. E. D.

I say next that *no other figure*, *besides the said five figures*, *can be constructed which is contained by equilateral and equiangular figures equal to one another.*

For a solid angle cannot be constructed with two triangles, or indeed planes.

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [[book-11/proposition-21|XI. 21]]

For the same reason, neither can a solid angle be constructed by more than six plane angles.

By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.

By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.

Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.

Therefore etc. Q. E. D.

Lemma. But that *the angle of the equilateral and equiangular pentagon is a right angle and a fifth* we must prove thus.

Let *ABCDE* be an equilateral and equiangular pentagon, let the circle *ABCDE* be circumscribed about it, let its centre *F* be taken, and let *FA*, *FB*, *FC*, *FD*, *FE* be joined.

Therefore they bisect the angles of the pentagon at *A*, *B*, *C*, *D*, *E*.

And, since the angles at *F* are equal to four right angles and are equal, therefore one of them, as the angle *AFB*, is one right angle less a fifth; therefore the remaining angles *FAB*, *ABF* consist of one right angle and a fifth.

But the angle *FAB* is equal to the angle *FBC*; therefore the whole angle *ABC* of the pentagon consists of one right angle and a fifth. Q. E. D.