Kitap 10
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Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.
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Let the square SQ be completed.
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It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the side of the area AC.
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It is now to be proved that MO is a first bimedial straight line.
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Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13]
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And, since AG is commensurable with EG, AE is also commensurable with each of the straight lines AG, GE. [X. 15]
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But AE is incommensurable in length with AB; therefore AG, GE are also incommensurable with AB. [X. 13]
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Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21]
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Hence each of the squares SN, NQ is medial.
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Therefore MN, NO are also medial.
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And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.
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And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11] that is, MN is incommensurable in length with NO.
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But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only.
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I say next that they also contain a rational rectangle.
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For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12]
