For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [[X. 75](elem.10.75)]

Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [[X. 15 and 23, Por.](elem.10.15 elem.10.23.p.1)]

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [[X. 22](elem.10.22)]

Now, since 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 LF is equal to twice the rectangle AG, GB. [[II. 7](elem.2.7)]

Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

But the rectangle AG, GB is medial; therefore FL is also medial.

And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [[X. 22](elem.10.22)]

And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [[VI. 1](elem.6.1), [X. 11](elem.10.11)]

But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB; therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. [[X. 13](elem.10.13)]

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.

But, as CL is to FL, so is CM to FM; [[VI. 1](elem.6.1)] therefore CM is incommensurable in length with FM. [[X. 11](elem.10.11)]

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [[X. 73](elem.10.73)]

I say next that it is also a third apotome.

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](elem.6.1), [X. 11](elem.10.11)]

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.

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [[VI. 1](elem.6.1)] therefore, as CK is to MN, so is MN to KM; [[V. 11](elem.5.11)] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM.

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](elem.10.17)]

And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome. [[X. Deff. III. 3](elem.10.def.3.3)]

Therefore etc. Q. E. D.