第10卷命题 108 · 有理面积减中项面积余边为两无理线之一

For from the rational area BC let the medial area BD be subtracted; I say that the “side” of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line. For let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to BC, and let GK equal to DB be subtracted; therefore the remainder EC is equal to LH. Since then BC is rational, and BD medial, while BC is equal to GH, and BD to GK, therefore GH is rational, and GK medial.

设有理面积BC减去中项面积BD，剩余EC，其边待证。

And they are applied to the rational straight line FG; therefore FH is rational and commensurable in length with FG, [X. 20] while FK is rational and incommensurable in length with FG; [X. 22] therefore FH is incommensurable in length with FK. [X. 13] Therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome [X. 73], and KF the annex to it. Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable.

取有理线段FG，作矩形GH等于BC，减去GK等于BD，则剩余EC等于LH。

First, let the square on it be greater by the square on a straight line commensurable with it. Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1] But the “side” of the rectangle contained by a rational straight line and a first apotome is an apotome.

因GH有理，GK中项，且均贴于FG，故FH与FG可公度，FK与FG不可公度，从而FH与FK仅平方可公度，KH为余线。

[X. 91] Therefore the “side” of LH, that is, of EC, is an apotome. But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, KH is a fourth apotome.

若HF上正方形大于FK上正方形之量可公度于HF，则KH为第一余线，其边为余线；若不可公度，则KH为第四余线，其边为次线。