The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.
第一双中项线段上的正方形，应用于一条有理线段，所产生的宽度是第二二项线。
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Let AB be a first bimedial straight line divided into its medials at C, of which medials 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 second binominal straight line. For let the same construction as before be made. Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21] Therefore DL is medial.
设AB为第一双中项线，被分为中项线AC和CB，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，宽度为DG。
[X. 15 and 23, Por.] And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22] Again, since twice the rectangle AC, CB is rational, MF is also rational. And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20] therefore DM is incommensurable in length with MG.
由于AB是第一双中项线，AC和CB是仅平方可通约的中项线，且它们所成矩形为有理，故AC和CB上的正方形也是中项面，从而DL是中项面，MD为有理且与DE长度不可通约。
[X. 13] And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] It is next to be proved that it is also a second binomial straight line. For, since the squares on AC, CB are greater than twice the rectangle AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG.
又因二倍矩形AC·CB为有理，MF为有理，MG与ML（即DE）长度可通约，故DM与MG长度不可通约，且均为有理，因此DM和MG是仅平方可通约的有理线段，DG为二项线。
[VI. 1] And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11] And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17] And MG is commensurable is length with DE.
由于AC和CB上的正方形之和大于二倍矩形AC·CB，故DL大于MF，DM大于MG；且AC上的正方形与CB上的正方形可通约，故DH与KL可通约，DK与KM可通约，从而DM上的正方形比MG上的正方形大一个与DM可通约的线段上的正方形，且MG与DE长度可通约，故DG是第二二项线。