Kitap 10
99
But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB; therefore CL is also incommensurable with FL.
99
But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]
99
And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]
99
I say next that it is also a third apotome.
99
For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11]
99
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.
99
But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM.
99
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 commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17]
