Let AB be an apotome, and BC an annex to it; therefore AC, CB are rational straight lines commensurable in square only. [X. 73] I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only. For, if possible, let BD be so annexed; therefore AD, DB are also rational straight lines commensurable in square only. [X. 73] Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB. But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational; therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26] Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only.
设AB是一个余线，BC是其附加线段；因此AC、CB是仅平方可公度的有理线段。我断言，不能再有另一条有理线段附加于AB，使得它与整个线段仅平方可公度。
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假设可能，设BD是另一条附加线段，则AD、DB也是仅平方可公度的有理线段。
由于AD、DB上的正方形之和减去二倍矩形AD、DB的差，等于AC、CB上的正方形之和减去二倍矩形AC、CB的差，两者都减去同一个量即AB上的正方形，因此交替地，AD、DB上的正方形之和减去AC、CB上的正方形之和的差，等于二倍矩形AD、DB减去二倍矩形AC、CB的差。
但AD、DB上的正方形之和减去AC、CB上的正方形之和的差是一个有理面积，因为两者都是有理的；因此二倍矩形AD、DB也减去二倍矩形AC、CB的差是一个有理面积。
这是不可能的，因为两者都是中项面积，而中项面积减去中项面积不能得到有理面积。因此不能再有另一条有理线段附加于AB，使得它与整个线段仅平方可公度。