第10卷命题 65 · 两中项和边产生第六二项线

Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, 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 sixth binomial straight line. For let the same construction be made as before. Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41] so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.

设AB为两中项和之边，在C点分割，DE为有理线段，在DE上作平行四边形DF等于AB上的正方形，DG为宽度。

And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22] And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore DL is incommensurable with MF.

由于AB是两中项和之边，AC、CB是平方不可通约的线段，它们的平方和是中项，它们所成矩形也是中项，且平方和与矩形不可通约，因此DL和MF都是中项。

Therefore DM is also incommensurable with MG; [VI. 1, X. 11] 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 sixth binomial straight line.

DL和MF应用于有理线段DE，故DM和MG都是有理线段，且长度上与DE不可通约。又因AC、CB的平方和与二倍矩形不可通约，故DL与MF不可通约，从而DM与MG不可通约，因此DM和MG是仅平方可通约的有理线段，故DG是二项线。

Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM. And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.

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，因此DM上的正方形比MG上的正方形大一个与DM长度不可通约的线段上的正方形，且DM和MG都与所设有理线段DE长度不可通约，故DG是第六二项线。