(related to Proposition: Prop. 10.095: Side of Area Contained by Rational Straight Line and Fifth Apotome)

- For let the area $AB$ have been contained by the rational (straight line) $AC$ and the fifth apotome $AD$.
- I say that the square root of area $AB$ is that (straight line) which with a rational (area) makes a medial whole.

- For let $DG$ be an attachment to $AD$.
- Thus, $AG$ and $DG$ are rational (straight lines which are) commensurable in square only [Prop. 10.73], and the attachment $GD$ is commensurable in length 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) incommensurable (in length) with ($AG$) [Def. 10.15] .
- Thus, if (some 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) incommensurable (in length) [Prop. 10.18].
- Therefore, let $DG$ have been divided in half at point $E$, and let (some 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 incommensurable in length with $FG$.
- And since $AG$ is incommensurable in length with $CA$, and both are rational (straight lines), $AK$ is thus a medial (area) [Prop. 10.21].
- Again, since $DG$ is rational, and commensurable in length with $AC$, $DK$ is a rational (area) [Prop. 10.19].
- Therefore, let the square $LM$, equal to $AI$, have been constructed.
- And let the square $NO$, equal to $FK$, (and) about the same angle, $LPM$, have been subtracted (from $NO$).
- 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.
- So, similarly (to the previous propositions), we can show that $LN$ is the square root of area $AB$.
- I say that $LN$ is that (straight line) which with a rational (area) makes a medial whole.
- For since $AK$ was shown (to be) a medial (area) , and is equal to (the sum of) the squares on $LP$ and $PN$, the sum of the (squares) on $LP$ and $PN$ is thus medial.
- Again, since $DK$ is rational, and is equal to twice the (rectangle contained) by $LP$ and $PN$, (the latter) is also rational.
- And since $AI$ is incommensurable with $FK$, the (square) on $LP$ is thus also incommensurable with the (square) on on $PN$.
- Thus, $LP$ and $PN$ are (straight lines which are) incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by them rational.
- Thus, the remainder $LN$ is the irrational (straight line) called that which with a rational (area) makes a medial whole [Prop. 10.77].
- And it is the square root of area $AB$.
- Thus, the square root of area $AB$ is that (straight line) which with a rational (area) makes a medial whole.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"