For again let the rectangle BD, G be equal to the square on A.
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]
But CB is greater than BD;
therefore G is also greater than EF. [V. 16, V. 14]
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]
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]
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.]
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]
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]
Therefore HK is commensurable in length with KE,
so that HE is also commensurable in length with EK. [X. 15]
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.
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.
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]
But KE is rational;
therefore FK is also rational.
Therefore FK, KE are rational straight lines commensurable in square only;
therefore EF is an apotome. [X. 73]