If an area be contained by a rational straight line and the second binomial, the “side” of the area is the irrational straight line which is called a first bimedial.
若一个面积由一条有理线段和第二条二项线围成，则该面积的“边”是一条无理线段，称为第一双中项线。
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For let the area ABCD be contained by the rational straight line AB and the second binomial AD; I say that the “side” of the area AC is a first bimedial straight line. For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term; therefore 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 the lesser term ED is commensurable in length with AB. [X. Deff. II. 2] Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17] Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, 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. Let the square SQ be completed.
设面积ABCD由有理线段AB和第二条二项线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上的正方形比ED上的正方形大一个与AE可通约的线段上的正方形，而较小项ED与AB长度可通约。
It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the “side” of the area AC. It is now to be proved that MO is a first bimedial straight line. Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13] And, since AG is commensurable with EG, AE is also commensurable with each of the straight lines AG, GE. [X. 15] But AE is incommensurable in length with AB; therefore AG, GE are also incommensurable with AB.
取ED的中点F，在AE上作矩形AG·GE等于EF上的正方形且缺一正方形，则AG与GE长度可通约。过G、E、F作GH、EK、FL平行于AB、CD，作正方形SN等于平行四边形AH，正方形NQ等于GK，并使MN与NO在一条直线上，则RN与NP也在一条直线上，完成正方形SQ。
[X. 13] Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21] Hence each of the squares SN, NQ is medial. Therefore MN, NO are also medial. And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO. And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11] that is, MN is incommensurable in length with NO.
由先前证明可知，MR是SN和NQ的比例中项且等于EL，而MO是面积AC的“边”。由于AE与ED长度不可通约，而ED与AB可通约，故AE与AB不可通约；又AG与GE可通约，故AE与AG、GE均可通约，从而AG、GE与AB不可通约，因此BA、AG和BA、GE是仅平方可通约的有理线段，故矩形AH和GK均为中项面，从而正方形SN和NQ均为中项面，故MN和NO也是中项线。
But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only. I say next that they also contain a rational rectangle. For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12] And each of them is rational; therefore EL, that is, MR is rational, [X. 19] and MR is the rectangle MN, NO. But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line.
由于AG与GE长度可通约，故AH与GK可通约，即SN与NQ可通约，亦即MN上的正方形与NO上的正方形可通约。又AE与ED长度不可通约，而AE与AG可通约，ED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，亦即PN与NR不可通约，故MN与NO长度不可通约。因此MN和NO是仅平方可通约的中项线，且它们所成的矩形为有理（因为DE与AB、EF均可通约，故EF与EK可通约，从而EL即MR为有理）。根据X.37，两仅平方可通约且含有理矩形的中项线之和为无理，称为第一双中项线。