To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.
求两条平方不可公度的线段，使得它们上的平方和是有理的，而它们所成的矩形是中项的。
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Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30] let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined. Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there has been applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE, EB, therefore AE is incommensurable with EB.
设AB、BC是仅平方可公度的有理线段，且AB上的正方形大于BC上的正方形，差为与AB不可公度的线段上的正方形。
[X. 18] And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB; therefore AF, FB are incommensurable in square. And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational.
取BC中点D，在AB上作等于BD或DC上的正方形的矩形，缺一正方形，得矩形AE、EB；以AB为直径作半圆AFB，作EF垂直于AB，连接AF、FB。
[I. 47] And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD; therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF. But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial.
由X.18，AE与EB不可公度；又AE比EB等于矩形BA、AE比AB、BE，而矩形BA、AE等于AF上的正方形，AB、BE等于BF上的正方形，故AF与FB平方不可公度。
[X. 23, Por.] But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma] therefore the rectangle AF, FB is also medial. But it was also proved that the sum of the squares on these straight lines is rational.
因AB有理，AB上的正方形有理，故AF、FB上的平方和有理；又矩形AE、EB等于EF上的正方形且等于BD上的正方形，故FE等于BD，BC是FE的二倍，从而矩形AB、BC与AB、EF可公度，而AB、BC是中项的，故AB、EF也是中项的，即矩形AF、FB是中项的。