# Proposition: Prop. 10.087: Construction of Third Apotome

### (Proposition 87 from Book 10 of Euclid's “Elements”)

To find a third apotome. * Let the rational (straight line) $A$ be laid down. * And let 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. * And let $CB$ 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 $FG$ [Prop. 10.6]. * And the square on $A$ (is) rational. * Thus, the (square) on $FG$ (is) also rational. * Thus, $FG$ is 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 a rational (straight line). * 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). * $FG$ and $GH$ are thus 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 third (apotome) .

### Modern Formulation

This proposition proves that the third apotome has length $\sqrt{\alpha}\,\left(1-\sqrt{1-\beta^{\,2}}\right),$

where $$\alpha,\beta$$ denote positive rational numbers.

### Notes

Proofs: 1

Propositions: 1

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