Kitap 10
101
I say next that it is also a fifth apotome.
101
For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM.
101
And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL.
101
But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]
101
Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]
101
And the annex FM is commensurable with the rational straight line CD set out; therefore CF is a fifth apotome. [X. Deff. III. 5] Q. E. D.
PROPOSITION 102.
102
The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome.
102
Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a sixth apotome.
