The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.
将一条有理线段应用于一条二项线段，所得宽度是一条余线段，其项与二项线段的项可公度且同比，且该余线段与二项线段同阶。
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Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A; I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC. For again let the rectangle BD, G be equal to the square on A. Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16] But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14] Let EH be equal to G; therefore, as CB is to BD, so is HE to EF; therefore, separando, as CD is to BD, so is HF to FE.
设A为有理线段，BC为二项线段，DC为较大项，矩形BC·EF等于A上的正方形。则EF是余线段，其项与CD、DB可公度且同比，且EF与BC同阶。
[V. 17] Let it be contrived that, as HF is to FE, so is FK to KE; therefore also the whole HK is to the whole KF as FK is to KE; for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12] But, as FK is to KE, so is CD to DB; [V. 11] therefore also, as HK is to KF, so is CD to DB. [id.] But the square on CD is commensurable with the square on DB; [X. 36] therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11] And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9] Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK.
作矩形BD·G等于A上的正方形。由BC·EF = BD·G得CB:BD = G:EF。因CB>BD，故G>EF。令EH=G，则CB:BD = HE:EF，分离得CD:BD = HF:FE。
[X. 15] Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational. And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD. Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11] But KE is rational; therefore FK is also rational. Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome.
设HF:FE = FK:KE，则HK:KF = FK:KE = CD:DB。因CD²与DB²可公度，故HK²与KF²可公度，从而HK与KE长度可公度，HE与EK长度可公度。
[X. 73] Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it. If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14] And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12] if BD is so commensurable, so also is KE; [X. 12] but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so. But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK.
因A² = EH·BD且A²为有理，故EH为有理且与BD长度可公度，从而EK也为有理且与BD长度可公度。由CD:DB = FK:KE且CD、DB仅平方可公度，得FK、KE仅平方可公度，故EF为余线段。若CD²比DB²大一个与CD可公度的线段上的正方形，则FK²比KE²大一个与FK可公度的线段上的正方形，且与有理线段的关系一致，故EF与BC同阶。