Book 10
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.
26
Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial.
26
And they are applied to the rational straight line EF; therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22]
