Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [[X. 6, Por.](elem.10.6.p.1)] therefore the square on EF is commensurable with the square on FG. [[X. 6](elem.10.6)]

Therefore FG is also rational.

Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number.

Therefore EF is incommensurable in length with FG; [[X. 9](elem.10.9)] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [[X. 36](elem.10.36)]

It is next to be proved that it is also a second binomial straight line.

For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE.

Let the squares on EF, H be equal to the square on GF; therefore, convertendo , as AB is to BC, so is the square on FG to the square on H. [[V. 19, Por.](elem.5.19.p.1)]

But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on H the ratio which a square number has to a square number.

Therefore FG is commensurable in length with H; [[X. 9](elem.10.9)] so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG.

And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out.

Therefore EG is a second binomial straight line. Q. E. D.