Kitap 10
25
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.
25
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]
25
And, since AD is commensurable with BE, therefore GH is also commensurable with NL.
25
And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11]
25
Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational. [X. 19]
25
And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO.
25
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.
25
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]
25
But the rectangle FH, KL is rational; therefore the square on HK is also rational.
25
Therefore HK is rational.
25
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]
25
Therefore HN is either rational or medial.
25
But HN is equal to AC; therefore AC is either rational or medial.
25
Therefore etc.
PROPOSITION 26.
26
4 medial area does not exceed a medial area by a rational area.
26
For, if possible, let the medial area AB exceed the medial area AC by the rational area DB, and let a rational straight line EF be set out; to EF let there be applied the rectangular parallelogram FH equal to AB, producing EH as breadth, and let the rectangle FG equal to AC be subtracted; therefore the remainder BD is equal to the remainder KH.
26
But DB is rational; therefore KH is also rational.
