The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.
若将一条中项线的正方形贴于一条有理线段上，则所得宽度为一条有理线段，且与所贴之线段长度不可公度。
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Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth; I say that CD is rational and incommensurable in length with CB. For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only. [X. 21] Let the square on it be equal to GF.
设A为中项线，CB为有理线，在BC上作矩形BD等于A上的正方形，得宽度CD。
But the square on it is also equal to BD; therefore BD is equal to GF. But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, proportionally, as BC is to EG, so is EF to CD. Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD.
因A为中项线，其正方形等于仅平方可公度的有理线段所成矩形（X.21），设该矩形为GF，则BD等于GF且等角，故对应边成反比（VI.14），得BC:EG = EF:CD。
[VI. 22] But the square on CB is commensurable with the square on EG, for each of these straight lines is rational; therefore the square on EF is also commensurable with the square on CD. [X. 11] But the square on EF is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is rational. And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG, [Lemma] therefore the square on EF is incommensurable with the rectangle FE, EG.
于是BC上的正方形与EG上的正方形之比等于EF上的正方形与CD上的正方形之比（VI.22）。因CB与EG均为有理线，其正方形可公度，故EF上的正方形与CD上的正方形可公度（X.11）。而EF上的正方形为有理，故CD上的正方形为有理（X.定义4），因此CD为有理线。
[X. 11] But the square on CD is commensurable with the square on EF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A; therefore the square on CD is also incommensurable with the rectangle DC, CB. [X. 13] But, as the square on CD is to the rectangle DC, CB, so is DC to CB; [Lemma] therefore DC is incommensurable in length with CB.
由于EF与EG仅平方可公度，长度不可公度，且EF:EG = EF上的正方形:矩形FE,EG（引理），故EF上的正方形与矩形FE,EG不可公度（X.11）。但CD上的正方形与EF上的正方形可公度，矩形DC,CB与矩形FE,EG相等（均等于A上的正方形），故CD上的正方形与矩形DC,CB不可公度（X.13）。而CD上的正方形:矩形DC,CB = DC:CB（引理），因此DC与CB长度不可公度（X.11）。