Proof: By Euclid
(related to Proposition: Prop. 10.093: Side of Area Contained by Rational Straight Line and Third Apotome)
- For let $DG$ be an attachment to $AD$.
- Thus, $AG$ and $GD$ are rational (straight lines which are) commensurable in square only [Prop. 10.73], and neither of $AG$ and $GD$ is commensurable in length with the (previously) laid down rational (straight line) $AC$, and the square on the whole, $AG$, is greater than (the square on) the attachment, $DG$, by the (square) on (some straight line) commensurable (in length) with ($AG$) [Def. 10.13] .
- Therefore, since the square on $AG$ is greater than (the square on) $GD$ by the (square) on (some straight line) commensurable (in length) with ($AG$), thus if (an area) equal to the fourth part of the square on $DG$ is applied to $AG$, falling short by a square figure, then it divides ($AG$) into (parts which are) commensurable (in length) [Prop. 10.17].
- Therefore, let $DG$ have been cut in half at $E$.
- And let (an area) equal to the (square) on $EG$ have been applied to $AG$, falling short by a square figure.
- And let it be the (rectangle contained) by $AF$ and $FG$.
- And let $EH$, $FI$, and $GK$ have been drawn through points $E$, $F$, and $G$ (respectively), parallel to $AC$.
- Thus, $AF$ and $FG$ are commensurable (in length) .
- $AI$ (is) thus also commensurable with $FK$ [Prop. 6.1], [Prop. 10.11].
- And since $AF$ and $FG$ are commensurable in length, $AG$ is thus also commensurable in length with each of $AF$ and $FG$ [Prop. 10.15].
- And $AG$ (is) rational, and incommensurable in length with $AC$.
- Hence, $AF$ and $FG$ (are) also (rational, and incommensurable in length with $AC$) [Prop. 10.13].
- Thus, $AI$ and $FK$ are each medial (areas) [Prop. 10.21].
- Again, since $DE$ is commensurable in length with $EG$, $DG$ is also commensurable in length with each of $DE$ and $EG$ [Prop. 10.15].
- And $GD$ (is) rational, and incommensurable in length with $AC$.
- Thus, $DE$ and $EG$ (are) each also rational, and incommensurable in length with $AC$ [Prop. 10.13].
- $DH$ and $EK$ are thus each medial (areas) [Prop. 10.21].
- And since $AG$ and $GD$ are commensurable in square only, $AG$ is thus incommensurable in length with $GD$.
- But, $AG$ is commensurable in length with $AF$, and $DG$ with $EG$.
- Thus, $AF$ is incommensurable in length with $EG$ [Prop. 10.13].
- And as $AF$ (is) to $EG$, so $AI$ is to $EK$ [Prop. 6.1].
- Thus, $AI$ is incommensurable with $EK$ [Prop. 10.11].
- Therefore, let the square $LM$, equal to $AI$, have been constructed.
- And let $NO$, equal to $FK$, which is about the same angle as $LM$, have been subtracted (from $LM$).
- Thus, $LM$ and $NO$ are about the same diagonal [Prop. 6.26].
- Let $PR$ be their (common) diagonal, and let the (rest of the) figure have been drawn.
- Therefore, since the (rectangle contained) by $AF$ and $FG$ is equal to the (square) on $EG$, thus as $AF$ is to $EG$, so $EG$ (is) to $FG$ [Prop. 6.17].
- But, as $AF$ (is) to $EG$, so $AI$ is to $EK$ [Prop. 6.1].
- And as $EG$ (is) to $FG$, so $EK$ is to $FK$ [Prop. 6.1].
- And thus as $AI$ (is) to $EK$, so $EK$ (is) to $FK$ [Prop. 5.11].
- Thus, $EK$ is in mean proportion3 to $AI$ and $FK$.
- And $MN$ is also in mean proportion3 to the squares $LM$ and $NO$ [Prop. 10.53 lem.] .
- And $AI$ is equal to $LM$, and $FK$ to $NO$.
- Thus, $EK$ is also equal to $MN$.
- But, $MN$ is equal to $LO$, and $EK$ [is] equal to $DH$ [Prop. 1.43].
- And thus the whole of $DK$ is equal to the gnomon $UVW$ and $NO$.
- And $AK$ (is) also equal to $LM$ and $NO$.
- Thus, the remainder $AB$ is equal to $ST$ - that is to say, to the square on $LN$.
- Thus, $LN$ is the square root of area $AB$.
- I say that $LN$ is the second apotome of a medial (straight line).
- For since $AI$ and $FK$ were shown (to be) medial (areas), and are equal to the (squares) on $LP$ and $PN$ (respectively), the (squares) on each of $LP$ and $PN$ (are) thus also medial.
- Thus, $LP$ and $PN$ (are) each medial (straight lines).
- And since $AI$ is commensurable with $FK$ [Prop. 6.1], [Prop. 10.11], the (square) on $LP$ (is) thus also commensurable with the (square) on on $PN$.
- Again, since $AI$ was shown (to be) incommensurable with $EK$, $LM$ is thus also incommensurable with $MN$ - that is to say, the (square) on $LP$ with the (rectangle contained) by $LP$ and $PN$.
- Hence, $LP$ is also incommensurable in length with $PN$ [Prop. 6.1], [Prop. 10.11].
- Thus, $LP$ and $PN$ are medial (straight lines which are) commensurable in square only.
- So, I say that they also contain a medial (area) .
- For since $EK$ was shown (to be) a medial (area) , and is equal to the (rectangle contained) by $LP$ and $PN$, the (rectangle contained) by $LP$ and $PN$ is thus also medial.
- Hence, $LP$ and $PN$ are medial (straight lines which are) commensurable in square only, and which contain a medial (area) .
- Thus, $LN$ is the second apotome of a medial (straight line) [Prop. 10.75].
- And it is the square root of area $AB$.
- Thus, the square root of area $AB$ is the second apotome of a medial (straight line).
- (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"
