Kitap 10
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Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36]
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It is next to be proved that it is also a sixth binomial straight line.
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For since, as D is to AB, so is the square on E to the square on FG, and also, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]
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But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]
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But it was also proved incommensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E.
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And, since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.
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Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]
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But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on FG to the square on K the ratio which a square number has to a square number.
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Therefore FG is incommensurable in length with K; [X. 9] therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG.
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And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out.
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Therefore FH is a sixth binomial straight line. Q. E. D.
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Lemma. Let there be two squares AB, BC, and let them be placed so that DB is in a straight line with BE; therefore FB is also in a straight line with BG.
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Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.
