Proof: By Euclid

Let the two numbers $AC$ and $CB$ be laid down such that $AB$ does not have to either of them the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. II] .
And let some rational straight line $D$ be laid down.
And let $EF$ be commensurable [in length] with $D$.
Thus, $EF$ (is) a rational (straight line).
And let it have been contrived that as $CA$ (is) to $AB$, so the (square) on $EF$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] .
And $CA$ does not have to $AB$ the ratio which (some) square number (has) to (some) square number.
Thus, the (square) on $EF$ does not have to the (square) on $FG$ the ratio which (some) square number (has) to (some) square number either.
Thus, $EF$ and $FG$ are rational (straight lines which are) commensurable in square only [Prop. 10.9].
Thus, $EG$ is a binomial (straight line) [Prop. 10.36].
So, I say that (it is) also a fifth (binomial straight line).

fig052e

For since as $CA$ is to $AB$, so the (square) on $EF$ (is) to the (square) on $FG$, inversely, as $BA$ (is) to $AC$, so the (square) on $FG$ (is) to the (square) on $FE$ [Prop. 5.7 corr.] 1.
Thus, the (square) on $GF$ (is) greater than the (square) on $FE$ [Prop. 5.14].
Therefore, let (the sum of) the (squares) on $EF$ and $H$ be equal to the (square) on $GF$.
Thus, via convertion, as the number $AB$ is to $BC$, so the (square) on $GF$ (is) to the (square) on $H$ [Prop. 5.19 corr.] 2.
And $AB$ does not have to $BC$ the ratio which (some) square number (has) to (some) square number.
Thus, the (square) on $FG$ does not have to the (square) on $H$ the ratio which (some) square number (has) to (some) square number either.
Thus, $FG$ is incommensurable in length with $H$ [Prop. 10.9].
Hence, the square on $FG$ is greater than (the square on) $FE$ by the (square) on (some straight line) incommensurable (in length) with ($FG$).
And $GF$ and $FE$ are rational (straight lines which are) commensurable in square only.
And the lesser term $EF$ is commensurable in length with the rational (straight line previously) laid down, $D$.
Thus, $EG$ is a fifth binomial (straight line).
(Which is) the very thing it was required to show.
∎

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