第10卷命题 92 · 第二类余线所成面积之边为第一中项线余线

For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the “side” of the area AB is a first apotome of a medial straight line. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2] Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let then 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 in length with FG. Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial.

设面积AB由有理线段AC和第二类余线AD围成，取AD的附加线段DG，则AG与GD是仅平方可通约的有理线段，且DG与AC长度可通约，而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。

[X. 21] Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15] But DG is commensurable in length with AC. Therefore each of the rectangles DH, EK is rational. [X. 19] Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM; 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 AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only.

在AG上作一个等于GD上正方形四分之一且缺一正方形图形的平行四边形，它分AG为可通约部分AF和FG，故AF与FG长度可通约，从而AG与AF、FG均长度可通约，但AG与AC长度不可通约，故AF、FG与AC长度不可通约，因此矩形AI和FK均为中项面。

And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11] But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore MN is also equal to EK. But DH is equal to EK, and LO equal to MN; therefore the whole DK is equal to the gnomon UVW and NO. Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS. But TS is the square on LN; therefore the square on LN is equal to the area AB; therefore LN is the “side” of the area AB. I say that LN is a first apotome of a medial straight line.

由于DE与EG可通约，故DG与DE、EG可通约，而DG与AC长度可通约，故矩形DH和EK均为有理面。作正方形LM等于AI，减去等于FK且与LM同角的正方形NO，则LM与NO共直径。因AI、FK为中项面，故LP、PN为中项线且仅平方可通约。又矩形AF、FG等于EG上的正方形，故EK是AI与FK的比例中项，而MN是LM与NO的比例中项，故MN等于EK，从而整个DK等于拐尺形UVW加NO，剩余AB等于TS，即LN上的正方形，故LN是面积AB的边。

For, since EK is rational and is equal to LO, therefore LO, that is, the rectangle LP, PN, is rational. But NO was proved medial; therefore LO is incommensurable with NO. But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP, PN are incommensurable in length. [X. 11] Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle; therefore LN is a first apotome of a medial straight line. [X. 74] And it is the “side” of the area AB.

由于EK为有理面且等于LO，故LO（即矩形LP、PN）为有理面，而NO为中项面，故LO与NO不可通约，从而LP与PN长度不可通约。因此LP、PN是仅平方可通约的中项线，且它们围成有理矩形，故LN是第一中项线余线，即为面积AB的边。