Proposition: Prop. 10.088: Construction of Fourth Apotome
(Proposition 88 from Book 10 of Euclid's “Elements”)
To find a 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) ]
Modern Formulation
This proposition proves that the fourth apotome has length
\[\alpha\frac \alpha{\sqrt{1+\beta}},\]
where \(\alpha,\beta\) denote positive rational numbers.
Notes
See also [Prop. 10.51].
Table of Contents
Proofs: 1
Mentioned in:
Propositions: 1
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"
Adapted from CC BYSA 3.0 Sources:
 Prime.mover and others: "Pr∞fWiki", https://proofwiki.org/wiki/Main_Page, 2016