The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational.
由两条仅平方可公度的有理线段所围成的矩形是无理的，且与该矩形面积相等的正方形的边也是无理的。后者称为中项线。
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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 the latter be called medial. For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB is equal to BD, therefore DB is also incommensurable in length with BC.
设矩形AC由两条仅平方可公度的有理线段AB、BC围成。
And, as DB is to BC, so is AD to AC; [VI. 1] therefore DA is incommensurable with AC. [X. 11] But DA is rational; therefore AC is irrational, so that the side of the square equal to AC is also irrational. [X. Def. 4] And let the latter be called medial. Q.
在AB上作正方形AD，则AD是有理的。由于AB与BC长度不可公度（仅平方可公度），且AB等于BD，故DB与BC长度不可公度。
E. D. LEMMA. If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.
由VI.1，DB比BC等于AD比AC，故由X.11，DA与AC不可公度。但DA是有理的，因此AC是无理的，从而与AC面积相等的正方形的边也是无理的。
Let FE, EG be two straight lines. I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG. For on FE let the square DF be described, and let GD be completed. Since then, as FE is to EG, so is FD to DG, [VI. 1] and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG.
称此边为中项线。引理：若有两条线段，则第一条比第二条等于第一条上的正方形比两线段所成矩形。