第10卷命题 100 · 小线段平方产生第四余线

Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fourth apotome. For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76] To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB. And the sum of the squares on AG, GB is rational; therefore CL is also rational. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20] And, 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 FL is equal to twice the rectangle AG, GB.

设AB为小线段，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。

[II. 7] Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the sum of the squares on AG, GB is rational, while twice the rectangle AG, GB is medial, the squares on AG, GB are incommensurable with twice the rectangle AG, GB. But CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB; therefore CL is incommensurable with FL.

取AB的附加线段BG，则AG、GB平方不可通约，且AG、GB平方和有理，而二倍矩形AG、GB中项。

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [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 that it is also a fourth apotome. For, since AG, GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB. And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL.

在CD上作CH等于AG上的正方形，宽度CK；KL等于GB上的正方形，宽度KM；则CL等于AG、GB平方和，为有理，故CM有理且与CD长度可通约。

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11] And, 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 square on GB to KL, and the rectangle AG, GB to NL, therefore NL is 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 [VI. 17], that is, to the fourth part of the square on FM. Since then 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 CM and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And the whole CM is commensurable in length with the rational straight line CD set out; therefore CF is a fourth apotome.

FL等于二倍矩形AG、GB，为中项，故FM有理且与CD长度不可通约；又CM与MF平方仅可通约，故CF为余线；且CK与KM长度不可通约，矩形CK、KM等于FM平方的四分之一，故CF为第四余线。