To a first 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 rational rectangle.
对于一条第一中项线段，只能附加一条中项线段，使得该线段与整体仅在平方上可公度，且与整体所成矩形为有理。
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For let AB be a first apotome of a medial straight line, and let BC be 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 rational; [X. 74] 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 rational area. For, if possible, let DB also be so annexed; therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational.
设AB为第一中项线段，BC为其附加线段，则AC、CB均为中项线段，仅在平方上可公度，且矩形AC、CB为有理。
[X. 74] 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 they 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 also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.
假设可附加另一线段DB，则AD、DB均为中项线段，仅在平方上可公度，且矩形AD、DB为有理。
But twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational. Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area.
由于AD、DB上的正方形和减去两倍矩形AD、DB的差等于AC、CB上的正方形和减去两倍矩形AC、CB的差，都等于AB上的正方形，因此两差相等。
which is impossible, for both are medial [X. 15 and 23, Por.], and a medial area does not exceed a medial by a rational area.
两倍矩形AD、DB减去两倍矩形AC、CB为有理面积，故AD、DB上的正方形和减去AC、CB上的正方形和也为有理面积，但中项面积之差不能为有理面积，矛盾。