◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 10.092: Side of Area Contained by Rational Straight Line and Second Apotome)

• For let the area $AB$ have been contained by the rational (straight line) $AC$ and the second apotome $AD$.
• I say that the square root of area $AB$ is the first apotome of a medial (straight line).

• 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 the attachment $DG$ 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, $GD$, by the (square) on (some straight line) commensurable in length with ($AG$) [Def. 10.12] .
• 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 $GD$ 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 $FG$.
• Thus, $AF$ is commensurable in length with $FG$.
• $AG$ is thus also commensurable in length with each of $AF$ and $FG$ [Prop. 10.15].
• And $AG$ (is) a rational (straight line), and incommensurable in length with $AC$.
• $AF$ and $FG$ are thus also each rational (straight lines), 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 $EG$, thus $DG$ is also commensurable (in length) with each of $DE$ and $EG$ [Prop. 10.15].
• But, $DG$ is commensurable in length with $AC$ [thus, $DE$ and $EG$ are also each rational, and commensurable in length with $AC$].
• Thus, $DH$ and $EK$ are each rational (areas) [Prop. 10.19].
• Therefore, let the square $LM$, equal to $AI$, have been constructed.
• And let $NO$, equal to $FK$, which is about the same angle $LPM$ as $LM$, have been subtracted (from $LM$).
• Thus, the squares $LM$ 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 $AI$ and $FK$ are medial (areas), and are equal to the (squares) on $LP$ and $PN$ (respectively), [thus] the (squares) on $LP$ and $PN$ are also medial.
• Thus, $LP$ and $PN$ are also medial (straight lines which are) commensurable in square only.¹
• And 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. 10.17].
• But, as $AF$ (is) to $EG$, so $AI$ (is) to $EK$.
• And as $EG$ (is) to $FG$, so $EK$ [is] to $FK$ [Prop. 6.1].
• Thus, $EK$ is in mean proportion3 to $AI$ and $FK$ [Prop. 5.11].
• 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, $MN$ is also equal to $EK$.
• But, $DH$ [is] equal to $EK$, and $LO$ equal to $MN$ [Prop. 1.43].
• Thus, the whole (of) $DK$ is equal to the gnomon $UVW$ and $NO$.
• Therefore, since the whole (of) $AK$ is equal to $LM$ and $NO$, of which $DK$ is equal to the gnomon $UVW$ and $NO$, the remainder $AB$ is thus equal to $TS$.
• And $TS$ is the (square) on $LN$.
• Thus, the (square) on $LN$ is equal to the area $AB$.
• $LN$ is thus the square root of area $AB$.
• So, I say that $LN$ is the first apotome of a medial (straight line).
• For since $EK$ is a rational (area), and is equal to $LO$, $LO$ - that is to say, the (rectangle contained) by $LP$ and $PN$ - is thus a rational (area).
• And $NO$ was shown (to be) a medial (area) .
• Thus, $LO$ is incommensurable with $NO$.
• And as $LO$ (is) to $NO$, so $LP$ is to $PN$ [Prop. 6.1].
• Thus, $LP$ and $PN$ are incommensurable in length [Prop. 10.11].
• $LP$ and $PN$ are thus medial (straight lines which are) commensurable in square only, and which contain a rational (area).
• Thus, $LN$ is the first apotome of a medial (straight line) [Prop. 10.74].
• And it is the square root of area $AB$.
• Thus, the square root of area $AB$ is the first 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)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"

Footnotes