Proof: By Euclid

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

fig049e

For since, inversely, as the number $BA$ is to $AC$, so the (square) on $GF$ (is) to the (square) on $FE$ [Prop. 5.7 corr.] 1, and $BA$ (is) greater than $AC$, the (square) on $GF$ (is) thus [also] greater than the (square) on $FE$ [Prop. 5.14].
Let (the sum of) the (squares) on $EF$ and $H$ be equal to the (square) on $GF$.
Thus, via convertion, as $AB$ is to $BC$, so the (square) on $FG$ (is) to the (square) on $H$ [Prop. 5.19 corr.] 2.
But, $AB$ has to $BC$ the ratio which (some) square number (has) to (some) square number.
Thus, the (square) on $FG$ also has to the (square) on $H$ the ratio which (some) square number (has) to (some) square number.
Thus, $FG$ is commensurable 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) commensurable in length with ($FG$).
And $FG$ 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) $D$ (previously) laid down.
Thus, $EG$ is a second binomial (straight line) [Def. 10.6] .
(Which is) the very thing it was required to show.
∎

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