Proof: By Euclid
(related to Proposition: Prop. 10.090: Construction of Sixth Apotome)
 Let the rational (straight line) $A$, and the three numbers $E$, $BC$, and $CD$, not having to one another the ratio which (some) square number (has) to (some) square number, be laid down.
 Furthermore, let $CB$ also not have to $BD$ the ratio which (some) square number (has) to (some) square number.
 And let it have been contrived that as $E$ (is) to $BC$, so the (square) on $A$ (is) to the (square) on $FG$, and as $BC$ (is) to $CD$, so the (square) on $FG$ (is) to the (square) on $GH$ [Prop. 10.6 corr.] .
 Therefore, since as $E$ is to $BC$, so the (square) on $A$ (is) to the (square) on $FG$, the (square) on $A$ (is) thus commensurable with the (square) on on $FG$ [Prop. 10.6].
 And the (square) on $A$ (is) rational.
 Thus, the (square) on $FG$ (is) also rational.
 Thus, $FG$ is also a rational (straight line).
 And since $E$ does not have to $BC$ the ratio which (some) square number (has) to (some) square number, the (square) on $A$ thus does not have to the (square) on $FG$ the ratio which (some) square number (has) to (some) square number either.
 Thus, $A$ is incommensurable in length with $FG$ [Prop. 10.9].
 Again, since as $BC$ is to $CD$, so the (square) on $FG$ (is) to the (square) on $GH$, the (square) on $FG$ (is) thus commensurable with the (square) on on $GH$ [Prop. 10.6].
 And the (square) on $FG$ (is) rational.
 Thus, the (square) on $GH$ (is) also rational.
 Thus, $GH$ (is) also rational.
 And since $BC$ does not have to $CD$ the ratio which (some) square number (has) to (some) square number, the (square) on $FG$ thus does not have to the (square) on $GH$ the ratio which (some) square (number) has to (some) square (number) either.
 Thus, $FG$ is incommensurable in length with $GH$ [Prop. 10.9].
 And both are rational (straight lines).
 Thus, $FG$ and $GH$ are rational (straight lines which are) commensurable in square only.
 Thus, $FH$ is an apotome [Prop. 10.73].

So, I say that (it is) also a sixth (apotome) .

For since as $E$ is to $BC$, so the (square) on $A$ (is) to the (square) on $FG$, and as $BC$ (is) to $CD$, so the (square) on $FG$ (is) to the (square) on $GH$, thus, via equality, as $E$ is to $CD$, so the (square) on $A$ (is) to the (square) on $GH$ [Prop. 5.22].
 And $E$ does not have to $CD$ the ratio which (some) square number (has) to (some) square number.
 Thus, the (square) on $A$ does not have to the (square) $GH$ the ratio which (some) square number (has) to (some) square number either.
 $A$ is thus incommensurable in length with $GH$ [Prop. 10.9].
 Thus, neither of $FG$ and $GH$ is commensurable in length with the rational (straight line) $A$.
 Therefore, let the (square) on $K$ be that (area) by which the (square) on $FG$ is greater than the (square) on $GH$ [Prop. 10.13 lem.] .
 Therefore, since as $BC$ is to $CD$, so the (square) on $FG$ (is) to the (square) on $GH$, thus, via convertion, as $CB$ is to $BD$, so the (square) on $FG$ (is) to the (square) on $K$ [Prop. 5.19 corr.] 2.
 And $CB$ does not have to $BD$ the ratio which (some) square number (has) to (some) square number.
 Thus, the (square) on $FG$ does not have to the (square) on $K$ the ratio which (some) square number (has) to (some) square number either.
 $FG$ is thus incommensurable in length with $K$ [Prop. 10.9].
 And the square on $FG$ is greater than (the square on) $GH$ by the (square) on $K$.
 Thus, the square on $FG$ is greater than (the square on) $GH$ by the (square) on (some straight line) incommensurable in length with ($FG$).
 And neither of $FG$ and $GH$ is commensurable in length with the (previously) laid down rational (straight line) $A$.
 Thus, $FH$ is a sixth apotome [Def. 10.16] .
 Thus, the sixth apotome $FH$ has been found.
 (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"