(related to Proposition: Prop. 10.108: Side of Remaining Area from Rational Area from which Medial Area Subtracted)

- For let the medial (area) $BD$ have been subtracted from the rational (area) $BC$.
- I say that one of two irrational (straight lines) arise (as) the square root of the remaining (area), $EC$ - either an apotome, or a minor (straight line).

- For let the rational (straight line) $FG$ have been laid out, and let the right-angled parallelogram $GH$, equal to $BC$, have been applied to $FG$, and let $GK$, equal to $DB$, have been subtracted (from $GH$).
- Thus, the remainder $EC$ is equal to $LH$.
- Therefore, since $BC$ is a rational (area), and $BD$ a medial (area) , and $BC$ (is) equal to $GH$, and $BD$ to $GK$, $GH$ is thus a rational (area), and $GK$ a medial (area) .
- And they are applied to the rational (straight line) $FG$.
- Thus, $FH$ (is) rational, and commensurable in length with $FG$ [Prop. 10.20], and $FK$ (is) also rational, and incommensurable in length with $FG$ [Prop. 10.22].
- Thus, $FH$ is incommensurable in length with $FK$ [Prop. 10.13].
- $FH$ and $FK$ are thus rational (straight lines which are) commensurable in square only.
- Thus, $KH$ is an apotome [Prop. 10.73], and $KF$ an attachment to it.
- So, the square on $HF$ is greater than (the square on) $FK$ by the (square) on (some straight line which is) either commensurable, or not (commensurable), (in length with $HF$).
- First, let the square (on it) be (greater) by the (square) on (some straight line which is) commensurable (in length with $HF$).
- And the whole of $HF$ is commensurable in length with the (previously) laid down rational (straight line) $FG$.
- Thus, $KH$ is a first apotome [Def. 10.1] .
- And the square root of an (area) contained by a rational (straight line) and a first apotome is an apotome [Prop. 10.91].
- Thus, the square root of $LH$ - that is to say, (of) $EC$ - is an apotome.
- And if the square on $HF$ is greater than (the square on) $FK$ by the (square) on (some straight line which is) incommensurable (in length) with ($HF$), and (since) the whole of $FH$ is commensurable in length with the (previously) laid down rational (straight line) $FG$, $KH$ is a fourth apotome [Prop. 10.14].
- And the square root of an (area) contained by a rational (straight line) and a fourth apotome is a minor (straight line) [Prop. 10.94].
- (Which is) the very thing it was required to show.∎

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