Kitap 13
18
But the side of the pentagon is the side of the icosahedron; [XIII. 16] therefore MB belongs to the icosahedron.
18
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. [XIII. 17, Por.]
18
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.
18
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.
18
The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.
18
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 [XIII. 16] and the other an apotome [XIII. 17].
18
That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus.
18
For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8] proportionally, as DB is to BF, so is BF to BA. [VI. 4]
18
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; [V. Def. 9, 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.
