Kitap 10
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Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30]
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Lemma. Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.
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And first that the rectangle CB, BD is equal to the square on BA.
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For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8]
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And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17]
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For the same reason the rectangle BC, CD is also equal to the square on AC.
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And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17]
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I say that the rectangle BC, AD is also equal to the rectangle BA, AC.
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For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4]
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Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16] Q. E. D.
PROPOSITION 33.
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To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.
