The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.
中项线上的正方形，应用于一条有理线段，所产生的宽度为第四二项线。
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Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line. Let the same construction be made as before shown. Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.
设AB为中项线，被C分为两段，AC大于CB；DE为有理线段，在DE上作平行四边形DF等于AB上的正方形，且DG为宽度。
[X. 39] Since then the sum of the squares on AC, CB is rational, therefore DL is rational; therefore DM is also rational and commensurable in length with DE. [X. 20] Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE; [X. 22] therefore DM is also incommensurable in length with MG. [X. 13] Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial.
由于AB是中项线，AC与CB是平方不可通约的线段，它们的平方和是有理的，而它们所成矩形是中项的，因此DL是有理的，DM也是有理的且与DE长度可通约。
[X. 36] It is to be proved that it is also a fourth binomial straight line. In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN. Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM.
又因为两倍矩形AC·CB（即MF）是中项的，且它应用于有理线段ML，所以MG是有理的但与DE长度不可通约，从而DM与MG长度不可通约，故DM与MG是仅平方可通约的有理线段，因此DG是二项线。
[VI. 1, X. 11] But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out.
类似前述可证DM大于MG，且矩形DK·KM等于MN上的正方形。由于AC上的正方形与CB上的正方形不可通约，故DH与KL不可通约，从而DK与KM不可通约。根据X.18，DM上的正方形大于MG上的正方形的部分等于一条与DM长度不可通约的线段上的正方形，且DM与给定有理线段DE可通约，因此DG是第四二项线。