Kitap 10
111
Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1]
111
Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60]
111
Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1]
111
Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15]
111
But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [X. 13]
111
Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73]
111
But it is also rational: which is impossible.
111
Therefore the apotome is not the same with the binomial straight line. Q. E. D.
PROPOSITION 112.
112
The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.
112
Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A; I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC.
112
For again let the rectangle BD, G be equal to the square on A.
