Kitap 10
102
For, since FL is equal to twice the rectangle AG, GB, let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.
102
And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB.
102
But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL.
102
But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [X. 11]
102
And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.
102
And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]
102
And neither of them is commensurable with the rational straight line CD set out; therefore CF is a sixth apotome. [X. Deff. III. 6] Q. E. D.
PROPOSITION 103.
103
A straight line commensurable in length with an apotome is an apotome and the same in order.
103
Let AB be an apotome, and let CD be commensurable in length with AB; I say that CD is also an apotome and the same in order with AB.
103
For, since AB is an apotome, let BE be the annex to it; therefore AE, EB are rational straight lines commensurable in square only. [X. 73]
103
Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12] therefore also, as one is to one, so are all to all; [V. 12] therefore also, as the whole AE is to the whole CF, so is AB to CD.
103
But AB is commensurable in length with CD.
103
Therefore AE is also commensurable with CF, and BE with DF. [X. 11]
103
And AE, EB are rational straight lines commensurable in square only; therefore CF, FD are also rational straight lines commensurable in square only. [X. 13]
