Proof: By Euclid

Let the two numbers $AC$ and $CB$ be laid down such that $AB$ does not have to $BC$, or to $AC$ either, 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 $D$.
Thus, $EF$ is also a rational (straight line).
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.] .
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 $BA$ does not have to $AC$ 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].
Thus, $EF$ and $FG$ are rational (straight lines which are) commensurable in square only.
Hence, $EG$ is a binomial (straight line) [Prop. 10.36].
So, I say that (it is) also a fourth (binomial straight line).

fig051e

For since as $BA$ 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 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 $EF$.
Thus, via convertion, as the number $AB$ (is) to $BC$, so the (square) on $EF$ (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 $EF$ does not have to the (square) on $H$ the ratio which (some) square number (has) to (some) square number either.
Thus, $EF$ is incommensurable in length with $H$ [Prop. 10.9].
Thus, the square on $EF$ is greater than (the square on) $GF$ by the (square) on (some straight line) incommensurable (in length) with ($EF$).
And $EF$ and $FG$ are rational (straight lines which are) commensurable in square only.
And $EF$ is commensurable in length with $D$.
Thus, $EG$ is a fourth binomial (straight line) [Def. 10.8] .
(Which is) the very thing it was required to show.
∎

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