A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order.
与中项线段的余线可公度的线段也是中项线段的余线，且阶次相同。
本页以“中项线段的余线可公度性”整体图解辅助阅读；点、线、角、圆索引已按命题文字和证明步骤校订，可与证明和问答联动。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB; I say that CD is also an apotome of a medial straight line and the same in order with AB. For, since AB is an apotome of a medial straight line, let EB be the annex to it. Therefore AE, EB are medial straight lines commensurable in square only.
设AB是中项线段的余线，EB是其附加线段，则AE、EB是仅平方可公度的中项线段。
[X. 74, 75] Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12] therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11] But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line.
作比例AB:CD = BE:DF，则AE与CF、BE与DF分别可公度。
[X. 74, 75] I say next that it is also the same in order with AB. 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.
由于AE、EB是仅平方可公度的中项线段，故CF、FD也是仅平方可公度的中项线段，因此CD是中项线段的余线。
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. [V. 16, X. 11] Therefore, if the rectangle AE, EB is rational, the rectangle CF, FD will also be rational, [X. Def. 4] and if the rectangle AE, EB is medial, the rectangle CF, FD is also medial.
由AE:EB = CF:FD，得AE上的正方形与AE、EB所成矩形之比等于CF上的正方形与CF、FD所成矩形之比。若前者为有理，则后者也有理；若前者为中项，则后者也为中项，故阶次相同。