Proof: By Euclid

Let the rational (straight line) $A$ be laid down.
And let $BG$ be commensurable in length with $A$.
$BG$ is thus also a rational (straight line).
And let two square numbers $DE$ and $EF$ be laid down, and let their difference $FD$ be not square [Prop. 10.28 lem. I] .
Thus, $ED$ does not have to $DF$ the ratio which (some) square number (has) to (some) square number.
And let it have been contrived that as $ED$ (is) to $DF$, so the square on $BG$ (is) to the square on $GC$ [Prop. 10.6 corr.] .
Thus, the (square) on $BG$ is commensurable with the (square) on on $GC$ [Prop. 10.6].
And the (square) on $BG$ (is) rational.
Thus, the (square) on $GC$ (is) also rational.
Thus, $GC$ is also rational.
And since $ED$ does not have to $DF$ the ratio which (some) square number (has) to (some) square number, the (square) on $BG$ thus does not have to the (square) on $GC$ the ratio which (some) square number (has) to (some) square number either.
Thus, $BG$ is incommensurable in length with $GC$ [Prop. 10.9].
And they are both rational (straight lines).
Thus, $BG$ and $GC$ are rational (straight lines which are) commensurable in square only.
Thus, $BC$ is an apotome [Prop. 10.73].
So, I say that (it is) also a first (apotome).

fig085e

Let the (square) on $H$ be that (area) by which the (square) on $BG$ is greater than the (square) on $GC$ [Prop. 10.13 lem.] .
And since as $ED$ is to $FD$, so the (square) on $BG$ (is) to the (square) on $GC$, thus, via convertion, as $DE$ is to $EF$, so the (square) on $GB$ (is) to the (square) on $H$ [Prop. 5.19 corr.] 2.
And $DE$ has to $EF$ the ratio which (some) square-number (has) to (some) square-number.
For each is a square (number) .
Thus, the (square) on $GB$ also has to the (square) on $H$ the ratio which (some) square number (has) to (some) square number.
Thus, $BG$ is commensurable in length with $H$ [Prop. 10.9].
And the square on $BG$ is greater than (the square on) $GC$ by the (square) on $H$.
Thus, the square on $BG$ is greater than (the square on) $GC$ by the (square) on (some straight line) commensurable in length with ($BG$).
And the whole, $BG$, is commensurable in length with the (previously) laid down rational (straight line) $A$.
Thus, $BC$ is a first apotome [Def. 10.11] .
Thus, the first apotome $BC$ has been found.
(Which is) the very thing it was required to find.
∎

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