Proof: By Euclid

Let 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 $CA$ the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I] .
And let some rational (straight line) $D$ be laid down.
And let $EF$ be commensurable in length with $D$.
$EF$ is thus also rational [Def. 10.3] .
And let it have been contrived that as the number $BA$ (is) to $AC$, so the (square) on $EF$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] .
And $AB$ has to $AC$ the ratio which (some) number (has) to (some) number.
Thus, the (square) on $EF$ also has to the (square) on $FG$ the ratio which (some) number (has) to (some) number.
Hence, the (square) on $EF$ is commensurable with the (square) on on $FG$ [Prop. 10.6].
And $EF$ is rational.
Thus, $FG$ (is) also rational.
And since $BA$ does not have to $AC$ 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, $EF$ 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].
I say that (it is) also a first (binomial straight line).

fig048e

For since as the number $BA$ is to $AC$, so the (square) on $EF$ (is) to the (square) on $FG$, and $BA$ (is) greater than $AC$, the (square) on $EF$ (is) thus also greater than the (square) on $FG$ [Prop. 5.14].
Therefore, let (the sum of) the (squares) on $FG$ and $H$ be equal to the (square) on $EF$.
And since as $BA$ is to $AC$, so the (square) on $EF$ (is) to the (square) on $FG$, thus, via convertion, as $AB$ is to $BC$, so the (square) on $EF$ (is) to the (square) on $H$ [Prop. 5.19 corr.] 2.
And $AB$ has to $BC$ the ratio which (some) square number (has) to (some) square number.
Thus, the (square) on $EF$ also has to the (square) on $H$ the ratio which (some) square number (has) to (some) square number.
Thus, $EF$ is commensurable in length with $H$ [Prop. 10.9].
Thus, the square on $EF$ is greater than (the square on) $FG$ by the (square) on (some straight line) commensurable (in length) with ($EF$).
And $EF$ and $FG$ are rational (straight lines).
And $EF$ (is) commensurable in length with $D$.
Thus, $EG$ is a first binomial (straight line) " [Def. 10.5] ":bookofproofs$2085.[^1]
(Which is) the very thing it was required to show.
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