第10卷命题 91 · 有理线段与第一余线所成面积之边为余线

For let the area AB be contained by the rational straight line AC and the first apotome AD; I say that the “side” of the area AB is an apotome. For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73] And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable with FG. And through the points E, F, G let EH, FI, GK be drawn parallel to AC. Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC.

设面积AB由有理线段AC和第一余线AD围成，取AD的附属线段DG，则AG和GD是仅平方可通约的有理线段。

[X. 12] And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19] Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15] But DG is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21] Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then the rectangle contained by AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG.

由于AG与AC可通约，且AG上的正方形比GD上的正方形大一个与AG长度可通约的线段上的正方形，故将平行四边形应用于AG，使其等于DG上正方形的四分之一且亏缺一个正方形，则它分成可通约部分。

[VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1] therefore EK is a mean proportional between AI, KF. [V. 11] But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO; therefore MN is also equal to EK. But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO. But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST. But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the “side” of AB. I say next that LN is an apotome.

取DG的中点E，作平行四边形AF·FG等于EG上的正方形且亏缺一个正方形，则AF与FG可通约。作平行线，证明AI和FK为有理矩形，DH和EK为中项矩形。

For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational. Again, since DH is medial and is equal to LO, therefore LO is also medial. Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO. But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP is incommensurable in length with PN. [X. 11] And both are rational; therefore LP, PN are rational straight lines commensurable in square only; therefore LN is an apotome. [X. 73] And it is the “side” of the area AB; therefore the “side” of the area AB is an apotome.

作正方形LM等于AI，减去正方形NO等于FK，则两正方形共对角线。通过比例关系证明LN上的正方形等于AB，且LP和PN是仅平方可通约的有理线段，故LN是余线。