Kitap 10
112
Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16]
112
But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14]
112
Let EH be equal to G; therefore, as CB is to BD, so is HE to EF; therefore, separando, as CD is to BD, so is HF to FE. [V. 17]
112
Let it be contrived that, as HF is to FE, so is FK to KE; therefore also the whole HK is to the whole KF as FK is to KE; for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12]
112
But, as FK is to KE, so is CD to DB; [V. 11] therefore also, as HK is to KF, so is CD to DB. [id.]
112
But the square on CD is commensurable with the square on DB; [X. 36] therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11]
112
And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9]
112
Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. [X. 15]
112
Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational.
112
And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD.
112
Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11]
112
But KE is rational; therefore FK is also rational.
112
Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome. [X. 73]
112
Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it.
