Okay, I found it. It's part of Book XIII, Prop, 18. Here I quote from Heath's translation:
"I say next that no other figure, besides the said five figures [tetrahedron, octahedron, cube, icosahedron, dodecahedron], 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.
"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 folur 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 slid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the fur 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."
He adds a lemma for which I will not show the proof:
"But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus."
[Proof omitted]
2006-08-28 08:53:06
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answer #1
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answered by bpiguy 7
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