To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.
求作两条线段，它们平方不可通约，且它们的平方和是中项线，但所成矩形是有理线。
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Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.] let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [X. 18] Let FD be drawn from F at right angles to AB, and let AD, DB be joined.
取两条中项线AB、BC，仅平方可通约，使它们所成矩形为有理，且AB上的正方形比BC上的正方形大出与AB不可通约的线段上的正方形。
Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11] But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB; therefore the square on AD is also incommensurable with the square on DB.
在AB上作半圆ADB，平分BC于E，在AB上作平行四边形等于BE上的正方形且缺一正方形，即矩形AF、FB，则AF与FB长度不可通约。
And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47] And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD.
从F作FD垂直于AB，连接AD、DB。由于AF与FB长度不可通约，故矩形BA、AF与AB、BF不可通约，而它们分别等于AD和DB上的正方形，因此AD与DB上的正方形不可通约。
But the rectangle AB, BC is rational; therefore the rectangle AB, FD is also rational. [X. 6] But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma] so that the rectangle AD, DB is also rational.
由于AB上的正方形是中项线，故AD、DB上的正方形之和也是中项线。又BC是DF的二倍，故矩形AB、BC是AB、FD的二倍，而AB、BC为有理，故AB、FD为有理，且等于AD、DB所成矩形，因此AD、DB所成矩形为有理。