The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.
将第二双中项线上的正方形应用于一条有理线段，其宽度为第三二项线。
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Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any 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 third binomial straight line. Let the same construction be made as before shown. Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38] so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.] And it is equal to DL; therefore DL is also medial.
设AB为第二双中项线，被分为中项线于C，其中AC为较大段；DE为任意有理线段，作平行四边形DF等于AB上的正方形，且DG为宽度。
And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. [X. 22] For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB, therefore the square on AC is also incommensurable with the rectangle AC, CB.
由于AB是第二双中项线，AC、CB是仅平方可通约且包含中项矩形的中项线，故AC、CB上的正方形之和也是中项的，等于DL，因此DL是中项的。
[X. 11] Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [X. 12, 13] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. 1, X. 11] And they are rational; therefore DG is binomial. [X. 36] It is to be proved that it is also a third binomial straight line.
DL应用于有理线段DE，故MD是有理的且与DE长度不可通约；同理MG是有理的且与DE长度不可通约。
In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM. 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. And neither of the straight lines DM, MG is commensurable in length with DE.
因AC与CB长度不可通约，故AC上的正方形与矩形AC、CB不可通约，从而DL与MF不可通约，故DM与MG不可通约；它们都是有理的，因此DG是二项线；进一步可证DM大于MG，且DK与KM可通约，故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DM、MG均与DE长度不可通约，因此DG是第三二项线。