Kitap 13
18
But AB is triple of BD; therefore the square on FB is triple of the square on BD.
18
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.
18
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; [XIII. 9] and, when FB is cut in extreme and mean ratio, NB is the greater segment; therefore KL is greater than NB.
18
But KL is equal to LM; therefore LM is greater than NB.
18
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.
18
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.
18
For a solid angle cannot be constructed with two triangles, or indeed planes.
18
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. [XI. 21]
18
For the same reason, neither can a solid angle be constructed by more than six plane angles.
18
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.
