第10卷命题 98 · 中项第一余线平方得第二余线

Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a second apotome. For let BG be the annex to AB;. therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74] To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD.

设AB是中项第一余线，CD是有理线段，在CD上作矩形CE等于AB上的正方形，宽度为CF。

[X. 22] Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7] But twice the rectangle AG, GB is rational; therefore FL is rational. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is also rational and commensurable in length with CD. [X. 20] Now, since the sum of the squares on AG, GB, that is, CL, is medial, while twice the rectangle AG, GB, that is, FL, is rational, therefore CL is incommensurable with FL.

取AB的附加线段BG，则AG、GB是仅平方可通约的中项线段，且它们所成矩形为有理。

But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a second apotome. For let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

在CD上作CH等于AG上的正方形，宽度为CK；作KL等于GB上的正方形，宽度为KM；则CL等于AG、GB上的正方形之和，为中线，其宽度CM为有理且与CD长度不可通约。

Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM. Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17] And the annex FM is commensurable in length with the rational straight line CD set out; therefore CF is a second apotome.

由于CL与FL（即二倍矩形AG·GB）不可通约，得CM与FM不可通约，故CF是余线；又因FM与CD可通约，且CM上的正方形大于MF上的正方形的部分等于与CM可通约的线段上的正方形，故CF是第二余线。