To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.
求作两条线段，它们平方不可公度，且其平方和为 medial，它们所成矩形也为 medial，且该矩形与平方和不可公度。
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Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32 , ad fin.] let the semicircle ADB be described on AB, and let the rest of the construction be as above. Then, since AF is incommensurable in length with FB, [X. 18 ] AD is also incommensurable in square with DB.
设两 medial 线段 AB、BC 仅平方可公度，所成矩形为 medial，且 AB 上的正方形比 BC 上的正方形大一个与 AB 长度不可公度的线段上的正方形。
[X. 11 ] 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 the rectangle AF, FB is equal to the square on each of the straight lines BE, DF, therefore BE is equal to DF; therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB, FD.
在 AB 上作半圆 ADB，其余构造同上。由于 AF 与 FB 长度不可公度，故 AD 与 DB 平方不可公度。
But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [X. 32, Por.] And it is equal to the rectangle AD, DB; [Lemma after X. 32 ] therefore the rectangle AD, DB is also medial.
因 AB 上的正方形为 medial，故 AD、DB 上的正方形之和也为 medial。又因矩形 AF、FB 等于 BE、DF 上的正方形，故 BE 等于 DF，从而 BC 是 FD 的二倍，因此矩形 AB、BC 是矩形 AB、FD 的二倍。
And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [X. 13 ] so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11 ] But the squares on AD, DB are equal to the square on AB, [I. 47 ] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.
矩形 AB、BC 为 medial，故矩形 AB、FD 也为 medial，且等于矩形 AD、DB，因此矩形 AD、DB 也为 medial。由于 AB 与 BC 长度不可公度，而 CB 与 BE 可公度，故 AB 与 BE 长度不可公度，从而 AB 上的正方形与矩形 AB、BE 不可公度。但 AD、DB 上的正方形和等于 AB 上的正方形，矩形 AB、FD 即矩形 AD、DB 等于矩形 AB、BE，因此 AD、DB 上的正方形和与矩形 AD、DB 不可公度。