Kitap 10
100
And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL, therefore NL is a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.
100
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 [VI. 17], that is, to the fourth part of the square on FM.
100
Since then CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF 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]
100
And the whole CM is commensurable in length with the rational straight line CD set out; therefore CF is a fourth apotome. [X. Deff. III. 4]
100
Therefore etc. Q. E. D.
PROPOSITION 101.
101
The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.
101
Let AB be the straight line which produces with a rational 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 fifth apotome.
101
For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77]
101
To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB.
101
But the sum of the squares on AG, GB together is medial; therefore CL is medial.
