For on AB, BC let the squares AD, BE be described;
therefore each of the squares AD, BE is medial.
Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth,
to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth,
and further to KN let there be similarly applied NL equal to BE, producing KL as breadth;
therefore FH, HK, KL are in a straight line.
Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL,
therefore each of the rectangles GH, NL is also medial.
And they are applied to the rational straight line FG;
therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22]
And, since AD is commensurable with BE,
therefore GH is also commensurable with NL.
And, as GH is to NL, so is FH to KL; [VI. 1]
therefore FH is commensurable in length with KL. [X. 11]
Therefore FH, KL are rational straight lines commensurable in length;
therefore the rectangle FH, KL is rational. [X. 19]
And, since DB is equal to BA, and OB to BC,
therefore, as DB is to BC, so is AB to BO.
But, as DB is to BC, so is DA to AC, [VI. 1]
and, as AB is to BO, so is AC to CO; [id.]
therefore, as DA is to AC, so is AC to CO.
But AD is equal to GH, AC to MK and CO to NL;
therefore, as GH is to MK, so is MK to NL;
therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11]
therefore the rectangle FH, KL is equal to the square on HK. [VI. 17]
But the rectangle FH, KL is rational;
therefore the square on HK is also rational.
Therefore HK is rational.
And, if it is commensurable in length with FG, HN is rational; [X. 19]
but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21]
Therefore HN is either rational or medial.
But HN is equal to AC;
therefore AC is either rational or medial.
Complete text
Book I