第10卷命题 67 · 双中项线可公度性

Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB. For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38] And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD.

设AB为双中项线，在E点分为两中项线，则AE、EB是仅平方可公度的中项线。

[V. 19] But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11] But AE, EB are medial; therefore CF, FD are also medial. [X. 23] And since, as AE is to EB, so is CF to FD, [V. 11] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only.

作比例AB:CD = AE:CF，则EB:FD = AB:CD，由AB与CD长度可公度得AE与CF、EB与FD分别可公度。

[X. 11] But they were also proved medial; therefore CD is bimedial. I say next that it is also the same in order with AB. For since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD; therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD.

因AE、EB是中项线，故CF、FD也是中项线；又由AE、EB仅平方可公度得CF、FD也仅平方可公度，所以CD是双中项线。

[V. 16] But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37] but if medial, medial, [X. 23, Por.] and each of the straight lines AB, CD is a second bimedial.

由AE:EB = CF:FD得AE平方:矩形AE,EB = CF平方:矩形CF,FD，交替后因AE平方与CF平方可公度，故矩形AE,EB与矩形CF,FD可公度；若前者为有理则后者也有理（第一双中项线），若前者为中项则后者也为中项（第二双中项线），故CD与AB同属一类。