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