第10卷命题 97 · 第十卷命题97：余线平方得第一余线

Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth; I say that CF is a first apotome. For let BG be the annex to AB; therefore AG, GB are rational straight lines commensurable in square only. [X. 73] To CD let there be applied CH equal to the square on AG, and KL equal to the square on BG. Therefore 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 FL is equal to twice the rectangle AG, GB. [II. 7] Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB. Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB,.

设AB为余线，CD为有理线段，在CD上作矩形CE等于AB上的正方形，宽度为CF。

therefore DM is rational. And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20] Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial. And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, 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为仅平方可通约的有理线段。在CD上作CH等于AG上的正方形，KL等于BG上的正方形，则CL等于AG、GB上的正方形之和，CE等于AB上的正方形，故FL等于二倍矩形AG·GB。

And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL. But, as DM 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 first apotome. For, 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 BG, 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.

取FM的中点N，过N作NO平行于CD，则FO、LN各等于矩形AG·GB。由于AG、GB上的正方形为有理，DM等于它们，故DM为有理，且CM与CD长度可通约。又二倍矩形AG·GB为中项线，FL等于它，故FL为中项线，FM与CD长度不可通约。

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 the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM. And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is commensurable with KM. [X. 11] Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, 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 CM is commensurable in length with the rational straight line CD set out; therefore CF is a first apotome.

由于AG、GB上的正方形与二倍矩形AG·GB不可通约，故DM与FL不可通约，从而CM与FM长度不可通约。CM、FM均为有理且仅平方可通约，故CF为余线。又因AG、GB上的正方形可通约，CH与KL可通约，故CK与KM可通约，且矩形CK·KM等于FM上正方形的四分之一，因此CM上的正方形大于MF上的正方形，其差等于与CM长度可通约的线段上的正方形，且CM与CD长度可通约，故CF为第一余线。