第10卷命题 54 · 第一二项线与有理线所容面积之边

For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the “side” of the area AC is the irrational straight line which is called binomial. For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term. It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1] Let ED be bisected at the point F. Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG.

设面积AC由有理线段AB和第一条二项线AD围成，将AD按项分为AE和ED，其中AE是较大项，则AE和ED是仅平方可通约的有理线段，且AE的平方大于ED的平方一个与AE可通约的线段上的平方，且AE与AB长度可通约。

Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP. And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma] Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17] therefore also, as AH is to EL, so is EL to KG; [VI. 1] therefore EL is a mean proportional between AH, GK. But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ. But MR is also a mean proportional between the same SN, NQ; [Lemma] therefore EL is equal to MR, so that it is also equal to PO.

取ED的中点F，在AE上作矩形AG·GE等于EF上的正方形，且为缺一正方形之形，则AG与EG长度可通约。作GH、EK、FL平行于AB或CD，构造正方形SN等于平行四边形AH，正方形NQ等于GK，并放置使MN与NO共线，完成正方形SQ。

But AH, GK are also equal to SN, NQ; therefore the whole AC is equal to the whole SQ, that is, to the square on MO; therefore MO is the “side” of AC. I say next that MO is binomial. For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15] But AE is also, by hypothesis, commensurable with AB; therefore AG, GE are also commensurable with AB. [X. 12] And AB is rational; therefore each of the straight lines AG, GE is also rational; therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK. But AH is equal to SN, and GK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable.

由AG·GE = EF²，得AG:EF = FE:EG，故AH:EL = EL:KG，即EL是AH和GK的比例中项。又AH = SN，GK = NQ，且MR也是SN和NQ的比例中项，故EL = MR = PO。而AH+GK = SN+NQ，故AC = SQ = MO²，因此MO是AC的边。

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, [X. 13] so that AH is also incommensurable with EL. [VI. 1, X. 11] But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR. But, as SN is to MR, so is PN to NR; [VI. 1] therefore PN is incommensurable with NR. [X. 11] But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO. And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only.

由于AG与GE可通约，故AE与AG、GE均可通约；又AE与AB可通约，故AG、GE与AB可通约，从而AH、GK为有理且可通约，即SN、NQ为有理且可通约。又AE与ED长度不可通约，而AE与AG可通约，DE与EF可通约，故AG与EF不可通约，从而AH与EL不可通约，即SN与MR不可通约，故PN与NR不可通约，即MN与NO不可通约。而MN²与NO²可通约且均为有理，故MN、NO是仅平方可通约的有理线段，即MO是二项线。