If an area be contained by a rational straight line and a fourth apotome, the “side” of the area is minor.
若一个面积由一条有理线段和一条第四余线围成，则该面积的“边”是次线。
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For let the area AB be contained by the rational straight line AC and the fourth apotome AD; I say that the “side” of the area AB is minor. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] 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 incommensurable in length with FG. Let EH, FI, GK be drawn through E, F, G parallel to AC, BD. Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19] Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial.
设面积AB由有理线段AC和第四余线AD围成，DG为AD的附加线，则AG与GD是仅平方可通约的有理线段，且AG与AC长度可通约，AG上的正方形比DG上的正方形大一个与AG长度不可通约的线段上的正方形。
[X. 21] Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11] Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, 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 the rectangle AF, FG is equal to the square on EG, therefore, proportionally, as AF is to EG, so is EG to FG.
在AG上作一个等于DG上正方形四分之一且缺一正方形的平行四边形，它分AG为不可通约的两段AF和FG。过E、F、G作AC、BD的平行线，得到矩形AI、FK等。
[VI. 17] But, 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 EK is also equal to MN. But DH is equal to EK, and LO is equal to MN; therefore the whole DK is equal to the gnomon UVW and NO. Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO, therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the “side” of the area AB. I say that LN is the irrational straight line called minor.
构造正方形LM等于AI，减去等于FK的正方形NO，两正方形共角且同直径。由于矩形AF·FG等于EG上的正方形，故EK是AI与FK的比例中项，也等于MN。因此整个DK等于拐尺形UVW加NO，从而剩余AB等于ST，即LN上的正方形。
For, since AK is rational and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is rational. Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial. And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN. Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. Therefore LN is the irrational straight line called minor; [X. 76] and it is the “side” of the area AB.
由于AK是有理的且等于LP、PN上的正方形之和，故该和有理；DK是中项线且等于二倍矩形LP·PN，故该二倍矩形中项；且LP上的正方形与PN上的正方形不可通约，因此LN是称为次线的无理线段。