Kitap 10
97
Therefore the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB; therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]
97
Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB.
97
Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB,. therefore DM is rational.
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And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20]
97
Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial.
97
And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]
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And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB.
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And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL.
97
But, as DM is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]
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And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]
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I say next that it is also a first apotome.
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For, 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 BG, 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.
97
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 the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.
