第10卷命题 110 · 第十卷命题110

For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole; I say that the “side” of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG.

设中项面BC减去与整体不可公度的中项面BD，则EC的边是两种无理线段之一。

[X. 22] And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11] therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome. [X. 73] If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome.

由于BC和BD都是中项面且不可公度，故FH和FK均为有理线段，且与FG长度不可公度。

[X. Deff. III. 3] But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the “side” of it is irrational, and is called a second apotome of a medial straight line; [X. 93] so that the “side” of LH, that is, of EC, is a second apotome of a medial straight line. But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF, FK is commensurable in length with FG, KH is a sixth apotome.

因BC与BD不可公度，即GH与GK不可公度，故HF与FK也不可公度，因此FH和FK是仅平方可公度的有理线段，故KH是余线。

[X. Deff. III. 6] But the “side” of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole.

若FH上的正方形大于FK上的正方形一个与FH可公度的线段上的正方形，且FH和FK均与有理线段FG长度不可公度，则KH是第三余线，此时LH的边是第二中项线段的余线；若差线段与FH不可公度，则KH是第六余线，此时LH的边是与中项面构成中项整体的线段。