第10卷命题 99 · 第二中项线段的端线产生第三端线

Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a third apotome. For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [X. 75] Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, 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. [X. 22] Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder LF is equal to twice the rectangle AG, GB.

设AB是第二中项线段的端线，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。

[II. 7] Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. But the rectangle AG, GB is medial; therefore FL is also medial. And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [X. 22] And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11] But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB; therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB.

取BG为AB的附加线段，则AG、GB是仅平方可通约且含中项矩形的中项线段。在CD上作CH等于AG上的正方形，宽度为CK；在KH上作KL等于GB上的正方形，宽度为KM；则整个CL等于AG、GB上的正方形之和，因此CL是中项线段。

[X. 13] But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB; therefore CL is also incommensurable with FL. 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 third apotome. For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM.

由于CL等于AG、GB上的正方形之和，而CE等于AB上的正方形，故剩余LF等于二倍的矩形AG、GB。将FM平分于N，作NO平行于CD，则FO和NL各等于矩形AG、GB，而矩形AG、GB是中项，故FL也是中项，因此FM是有理线段且与CD长度不可通约。

[VI. 1, X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, 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 KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM. Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to 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 with CM. [X. 17] And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome.

由于AG、GB仅平方可通约，故AG与GB长度不可通约，从而AG上的正方形与矩形AG、GB不可通约，进而CL与FL不可通约，因此CM与FM长度不可通约。又两者均为有理线段，故CM、MF是仅平方可通约的有理线段，所以CF是端线。又因CH与KL可通约，故CK与KM可通约，且NL是CH、KL的比例中项，从而CK、KM上的矩形等于FM上的正方形的四分之一，且CM、MF不等，故CM上的正方形比MF上的正方形大一个与CM可通约的线段上的正方形，且CM、MF均与CD长度不可通约，因此CF是第三端线。