To find medial straight lines commensurable in square only which contain a rational rectangle.
求作两条仅平方可通约的中项线段，使它们所成矩形为有理。
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Let two rational straight lines A, B commensurable in square only be set out; let C be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as A is to B, so is C to D. [VI. 12] Then, since A, B are rational and commensurable in square only, the rectangle A, B, that is, the square on C [VI.17], is medial. [X. 21] Therefore C is medial.
设两条仅平方可通约的有理线段A、B，取C为A、B的比例中项，并作A比B等于C比D。
[X. 21] And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [X. 11] And C is medial; therefore D is also medial.
由于A、B为有理且仅平方可通约，矩形A、B即C上的正方形为中项，故C为中项。
[X. 23, addition] Therefore C, D are medial and commensurable in square only. I say that they also contain a rational rectangle. For since, as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D.
因A比B等于C比D，且A、B仅平方可通约，故C、D也仅平方可通约；C为中项，故D也为中项。
[V. 16] But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B. But the square on B is rational; therefore the rectangle C, D is also rational.
由A比C等于C比B及A比C等于B比D，得C比B等于B比D，故矩形C、D等于B上的正方形，后者为有理，故矩形C、D为有理。