Proof: By Euclid
(related to Proposition: Prop. 10.088: Construction of Fourth Apotome)
 Let the rational (straight line) $A$, and $BG$ (which is) commensurable in length with $A$, be laid down.
 Thus, $BG$ is also a rational (straight line).
 And let the two numbers $DF$ and $FE$ be laid down such that the whole, $DE$, does not have to each of $DF$ and $EF$ the ratio which (some) square number (has) to (some) square number.
 And let it have been contrived that as $DE$ (is) to $EF$, so the square on $BG$ (is) to the (square) on $GC$ [Prop. 10.6 corr.] .
 The (square) on $BG$ is thus 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) a rational (straight line).
 And since $DE$ does not have to $EF$ 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 fourth (apotome) .

Now, 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.] .
 Therefore, since as $DE$ is to $EF$, so the (square) on $BG$ (is) to the (square) on $GC$, thus, also, via convertion, as $ED$ is to $DF$, so the (square) on $GB$ (is) to the (square) on $H$ [Prop. 5.19 corr.] 2.
 And $ED$ does not have to $DF$ the ratio which (some) square number (has) to (some) square number.
 Thus, the (square) on $GB$ does not have to the (square) on $H$ the ratio which (some) square number (has) to (some) square number either.
 Thus, $BG$ is incommensurable 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) incommensurable (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 fourth apotome [Def. 10.14] .
 Thus, a fourth apotome has been found.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"