Proof: By Euclid
(related to Proposition: Prop. 10.089: Construction of Fifth Apotome)
 Let the rational (straight line) $A$ be laid down, and let $CG$ be commensurable in length with $A$.
 Thus, $CG$ [is] a rational (straight line).
 And let the two numbers $DF$ and $FE$ be laid down such that $DE$ again does not have to each of $DF$ and $FE$ the ratio which (some) square number (has) to (some) square number.
 And let it have been contrived that as $FE$ (is) to $ED$, so the (square) on $CG$ (is) to the (square) on $GB$.
 Thus, the (square) on $GB$ (is) also rational [Prop. 10.6].
 Thus, $BG$ is also rational.
 And since as $DE$ is to $EF$, so the (square) on $BG$ (is) to the (square) on $GC$.
 And $DE$ 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).
 $BG$ and $GC$ are thus 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 fifth (apotome) .

For, 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 the (square) on $BG$ (is) to the (square) on $GC$, so $DE$ (is) to $EF$, thus, via convertion, as $ED$ is to $DF$, so the (square) on $BG$ (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 $BG$ 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 $GB$ is greater than (the square on) $GC$ by the (square) on (some straight line) incommensurable in length with ($GB$).
 And the attachment $CG$ is commensurable in length with the (previously) laid down rational (straight line) $A$.
 Thus, $BC$ is a fifth apotome [Def. 10.15] .
 Thus, the fifth apotome $BC$ 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"