A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.
与能产生中面中全线的线段可公度的线段，其本身也是能产生中面中全线的线段。
本页以“与中面中全线可公度线段亦为中面中全线”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 straight line which produces with a medial area a medial whole, and let CD be commensurable with AB; I say that CD is also a straight line which produces with a medial area a medial whole.
设AB为能产生中面中全线的线段，CD与AB可公度。
For let BE be the annex to AB, and let the same construction be made; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them.
取BE为AB的附加线段，作相同构造；则AE、EB是平方不可公度的线段，它们平方和为中面，所成矩形为中面，且平方和与矩形不可公度。
[X. 78] Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them.
已证AE、EB分别与CF、FD可公度，AE、EB的平方和与CF、FD的平方和可公度，矩形AE、EB与矩形CF、FD可公度。
因此CF、FD也是平方不可公度的线段，它们平方和为中面，所成矩形为中面，且平方和与矩形不可公度。