To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.
对于一条第二中项线段的第二余线，只能附加一条中项线段，该线段与整体仅平方可通约，且与整体所成矩形为中项矩形。
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Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [X. 75] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle. For, if possible, let BD also be so annexed; therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [X. 75] Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be subtracted, producing HM as breadth; therefore the remainder EL is equal to the square on AB, [II. 7] so that AB is the “side” of EL. Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.
设AB为第二中项线段的第二余线，BC为附加线段，则AC、CB均为中项线段，仅平方可通约，且所成矩形AC·CB为中项矩形。
But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7] Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial. And they are equal to EG; therefore EG is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22] Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial.
假设还可附加另一条中项线段BD，则AD、DB也为中项线段，仅平方可通约，且所成矩形AD·DB为中项矩形。
[X. 23, Por.] And it is equal to HG; therefore HG is also medial. And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is also rational and incommensurable in length with EF. [X. 22] And, since AC, CB are commensurable in square only, therefore AC is incommensurable in length with CB. But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11] But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6] therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB.
作有理线段EF，在其上作矩形EG等于AC、CB上的正方形和，得宽EM；减去HG等于二倍矩形AC·CB，得宽HM；则余量EL等于AB上的正方形，故AB为EL的边。同样，作EI等于AD、DB上的正方形和，得宽EN；则余量HI等于二倍矩形AD·DB。
[X. 13] And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HG. But, as EG is to HG, so is EM to HM; [VI. 1] therefore EM is incommensurable in length with MH. [X. 11] And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73] Similarly we can prove that HN is also an annex to it; therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible.
由于AC、CB仅平方可通约，故AC与CB长度不可通约，从而AC上的正方形与矩形AC·CB不可通约，进而EG与HG不可通约，因此EM与HM仅平方可通约，故EH为余线，HM为其附加线段。同理可证HN也为附加线段，这与一条余线只能附加一条仅平方可通约的线段矛盾。