Kitap 13
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For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB.
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I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid
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For, since AC is double of CB, therefore AB is triple of BC; and, convertendo, BA is one and a half times AC.
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But, as BA is to AC, so is the square on BA to the square on AD.
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Therefore the square on BA is also one and a half times the square on AD.
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And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid.
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Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.
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Lemma. It is to be proved that, as AB is to BC, so is the square on AD to the square on DC.
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For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.
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Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]
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And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB, therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.
