Kitap 10
54
For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the side of the area AC is the irrational straight line which is called binomial.
54
For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.
54
It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]
54
Let ED be bisected at the point F.
54
Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17]
54
Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG.
54
Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.
54
And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma]
