Kitap 10
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Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma] and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC.
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For let BC be bisected at the point E, and let EF be made equal to DE.
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Therefore the remainder DC is equal to BF.
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And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5]
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And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC.
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But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE.
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And the square on BC is equal to four times the square on EC, for again BC is double of CE.
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Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF.
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It is to be proved that BC is also commensurable with DF.
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Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15]
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But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6]
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Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.
