The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.
将有理中面线之边上的正方形应用于有理线段，所得宽度为第五二项线。
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Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a fifth binomial straight line. Let the same construction as before be made. Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.
设AB为有理中面线之边，被分为AC、CB两段，其中AC较大；取有理线段DE，作平行四边形DF等于AB上的正方形，得宽度DG。
[X. 40] Since then the sum of the squares on AC, CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. [X. 22] Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE.
由X.40，AC、CB为平方不可通约线段，其平方和为中面，所成矩形为有理。故DL为中面，DM为有理且与DE长度不可通约。
[X. 20] Therefore DM is incommensurable with MG; [X. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] I say next that it is also a fifth binomial straight line.
又由X.20，两倍矩形AC·CB即MF为有理，故MG为有理且与DE可通约。由X.13，DM与MG不可通约，故DM、MG为仅平方可通约的有理线段，因此DG为二项线。
For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18] And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE.
类似可证矩形DK·KM等于MN上的正方形，且DK与KM长度不可通约，故DM上的正方形大于MG上的正方形之部分与DM不可通约。且DM、MG仅平方可通约，较小者MG与DE长度可通约，故DG为第五二项线。