A straight line commensurable with a minor straight line is minor.
与一条次线段可公度的线段也是次线段。
本页以“与次线段可公度的线段是次线段”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 a minor straight line, and CD commensurable with AB; I say that CD is also minor. Let the same construction be made as before; then, since AE, EB are incommensurable in square, [X. 76] therefore CF, FD are also incommensurable in square.
设AB是次线段，CD与AB可公度，则作与之前相同的构造。
[X. 13] Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22] Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD.
由于AE、EB在平方上不可公度，故CF、FD也在平方上不可公度。
[V. 18] But the square on BE is commensurable with the square on DF; therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11] But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational.
由AE:EB = CF:FD，得平方比相等，合比后平方和与平方的比也相等；又BE平方与DF平方可公度，故AE、EB的平方和与CF、FD的平方和可公度，而前者为有理，故后者也为有理。
[X. Def. 4] Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. But the rectangle AE, EB is medial; [X. 76] therefore the rectangle CF, FD is also medial; [X. 23, Por.] therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.
又由AE平方与矩形AE·EB的比等于CF平方与矩形CF·FD的比，且AE平方与CF平方可公度，故矩形AE·EB与矩形CF·FD可公度；但前者为中项线，故后者也为中项线。因此CF、FD是平方不可公度、平方和为有理、所成矩形为中项线的两线段，故CD是次线段。