The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome.
与中面合成中面整体的线段上的正方形，若应用于一条有理线段，则其宽度为第六种余线。
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Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a sixth 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 them medial, twice the rectangle AG, GB medial, and the squares on AG, GB incommensurable with twice the rectangle AG, GB. [X. 78] Now 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; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. 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] Since now 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为与中面合成中面整体的线段，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] And twice the rectangle AG, GB is medial; 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 squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF.
取AB的附加段BG，则AG、GB平方不可通约，且两平方和为中面，两倍矩形AG·GB也为中面，且两平方和与两倍矩形不可通约。
[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 sixth apotome. For, since FL is equal to twice the rectangle AG, GB, 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. And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB.
在CD上作CH等于AG上的正方形，宽度为CK，及KL等于BG上的正方形，则CL等于两平方和，故CL为中面，其宽度CM为有理且与CD长度不可通约。
But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [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. And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM.
由于CE等于AB上的正方形，故FL等于两倍矩形AG·GB，也为中面，其宽度FM为有理且与CD长度不可通约；又CL与FL不可通约，故CM与MF长度不可通约，且两者均为有理，故CF为余线；进一步，因AG、GB平方不可通约，CH与KL不可通约，故CK与KM不可通约，且NL为CH、KL的比例中项，从而CM上的正方形大于MF上的正方形，其差为与CM不可通约线段上的正方形，故CF为第六种余线。