To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.
对于一条与中面构成中面整体的线段，只能附加一条线段，该线段与整条线段平方不可通约，且与整条线段一起使得两线段上的平方和为中面，两线段所成矩形的二倍也为中面，并且与两线段上的平方和不可通约。
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Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78] I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions. For, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [X. 78] 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 applied to EF, producing HM as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the “side” of EL.
设AB为与中面构成中面整体的线段，BC为其附线，则AC、CB是平方不可通约的线段，满足前述条件。
Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth. But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI. Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial. And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF.
假设可附加另一线段BD，使得AD、DB也是平方不可通约的线段，且满足类似条件。作有理线段EF，在其上作矩形EG等于AC、CB上的平方和，宽为EM；作矩形HG等于二倍矩形AC、CB，宽为HM；则剩余部分为AB上的正方形，即EL。
[X. 22] Again, since twice the rectangle AC, CB is medial and 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 rational and incommensurable in length with EF. [X. 22] And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB, EG is also incommensurable with HG; therefore EM is also incommensurable in length with MH. [VI. 1, 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.
同样，作矩形EI等于AD、DB上的平方和，宽为EN；则剩余部分HI等于二倍矩形AD、DB。由于EG和HG均为中面，故EM和HM为有理线段且与EF长度不可通约；又因EG与HG不可通约，故EM与MH长度不可通约，从而EH为余线，HM为其附线。
[X. 73] Similarly we can prove that EH is again an apotome and HN an annex to it. Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only: which was proved impossible. [X. 79] Therefore no other straight line can be so annexed to AB.
同理可证EH也是余线，HN为其附线。这样，同一余线有不同的有理附线，且与整条线段仅平方可通约，这与已知结论矛盾。因此，AB只能附加一条这样的线段。