Book 13
18
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.
18
Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.
18
Therefore etc. Q. E. D.
18
Lemma. But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.
18
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.
18
Therefore they bisect the angles of the pentagon at A, B, C, D, E.
18
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.
18
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.
