If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater.
设有两条不相等的线段，若将较大线段上作一个平行四边形，使其等于较小线段上的正方形的四分之一，且缺失一个正方形，且该平行四边形将较大线段分成不可公度的两部分，则较大线段上的正方形比较小线段上的正方形大一个线段上的正方形，该线段与较大线段不可公度。反之，若较大线段上的正方形比较小线段上的正方形大一个线段上的正方形，该线段与较大线段不可公度，且将较大线段上作一个平行四边形，使其等于较小线段上的正方形的四分之一，且缺失一个正方形，则该平行四边形将较大线段分成不可公度的两部分。
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And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable. Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD. It is to be proved that BC is incommensurable in length with DF.
设A、BC为两条不相等的线段，BC较大。在BC上作平行四边形等于A上的正方形的四分之一，且缺失一个正方形，即矩形BD、DC，且BD与DC长度不可公度。
Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16] But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16] And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure.
仿前作图，可证BC上的正方形比A上的正方形大FD上的正方形。需证BC与DF长度不可公度。
Let this be the rectangle BD, DC. It is to be proved that BD is incommensurable in length with DC. For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD. so that BC is also incommensurable with the remainder, the sum of BF, DC.
因BD与DC长度不可公度，故BC与CD亦不可公度（X.16）。但DC与BF、DC之和可公度（X.6），故BC与BF、DC之和不可公度（X.13），从而BC与余量FD长度不可公度（X.16）。
[X. 16] But the sum of BF, DC is commensurable in length with DC; [X. 6] therefore BC is also incommensurable in length with DC, [X. 13] so that, separando, BD is also incommensurable in length with DC. [X. 16] Therefore etc. LEMMA. [Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line be commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also.
反之，设BC上的正方形比A上的正方形大与BC不可公度的线段上的正方形，且在BC上作平行四边形等于A上的正方形的四分之一且缺失正方形，即矩形BD、DC。仿前可证BC上的正方形比A上的正方形大FD上的正方形。因BC与FD不可公度，故BC与BF、DC之和不可公度（X.16），而BF、DC之和与DC可公度（X.6），故BC与DC不可公度（X.13），从而BD与DC亦不可公度（X.16）。