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) ]
This proposition proves that the fourth apotome has length \[\alpha-\frac \alpha{\sqrt{1+\beta}},\]
where \(\alpha,\beta\) denote positive rational numbers.
See also [Prop. 10.51].
Proofs: 1
Propositions: 1
