To find two rational straight lines commensurable in square only and such that the square on the greater is greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater.
求两条仅平方可公度的有理线段，使得较大线段上的正方形比较小线段上的正方形大出一个与较大线段长度不可公度的线段上的正方形。
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Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2] let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6, Por.] and let FB be joined. Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only.
设AB为一条有理线段，取两个平方数CE和ED，使它们的和CD不是平方数。
And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF.
在AB上作半圆AFB，并作比DC:CE = 正方形BA²:正方形AF²，连接FB。
[V. 19, Por., III. 31, I. 47] But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF.
类似前文可证BA、AF是仅平方可公度的有理线段。
[X. 9] And the square on AB is greater than the square on AF by the square on FB incommensurable with AB.
由比例转换得CD:DE = 正方形AB²:正方形BF²，但CD与DE之比不是平方数比，故AB与BF长度不可公度，且正方形AB²比正方形AF²大出与AB不可公度的FB上的正方形。