(related to Proposition: Prop. 10.044: Second Bimedial Straight Line is Divisible Uniquely)

- Let $AB$ be a second bimedial (straight line) which has been divided at $C$, so that $AC$ and $BC$ are medial (straight lines), commensurable in square only, (and) containing a medial (area) [Prop. 10.38].
- So, (it is) clear that $C$ is not (located) at the point of bisection, since ($AC$ and $BC$) are not commensurable in length.
- I say that $AB$ cannot be (so) divided at another point.

- For, if possible, let it also have been (so) divided at $D$, so that $AC$ is not the same as $DB$, but $AC$ (is), by hypothesis, greater.
- So, (it is) clear that (the sum of) the (squares) on $AD$ and $DB$ is also less than (the sum of) the (squares) on $AC$ and $CB$, as we showed above [Prop. 10.41 lem.] .
- And $AD$ and $DB$ are medial (straight lines), commensurable in square only, (and) containing a medial (area) .
- And let the rational (straight line) $EF$ be laid down.
- And let the rectangular parallelogram $EK$, equal to the (square) on $AB$, have been applied to $EF$.
- And let $EG$, equal to (the sum of) the (squares) on $AC$ and $CB$, have been cut off (from $EK$).
- Thus, the remainder, $HK$, is equal to twice the (rectangle contained) by $AC$ and $CB$ [Prop. 2.4].
- So, again, let $EL$, equal to (the sum of) the (squares) on $AD$ and $DB$ - which was shown (to be) less than (the sum of) the (squares) on $AC$ and $CB$ - have been cut off (from $EK$).
- And, thus, the remainder, $MK$, (is) equal to twice the (rectangle contained) by $AD$ and $DB$.
- And since (the sum of) the (squares) on $AC$ and $CB$ is medial, $EG$ (is) thus [also] [medial]bookofproofs$2115.
- And it is applied to the rational (straight line) $EF$.
- Thus, $EH$ is rational, and incommensurable in length with $EF$ [Prop. 10.22].
- So, for the same (reasons), $HN$ is also rational, and incommensurable in length with $EF$.
- And since $AC$ and $CB$ are medial (straight lines which are) commensurable in square only, $AC$ is thus incommensurable in length with $CB$.
- And as $AC$ (is) to $CB$, so the (square) on $AC$ (is) to the (rectangle contained) by $AC$ and $CB$ [Prop. 10.21 lem.] .
- Thus, the (square) on $AC$ is incommensurable with the (rectangle contained) by $AC$ and $CB$ [Prop. 10.11].
- But, (the sum of) the (squares) on $AC$ and $CB$ is commensurable with the (square) on on $AC$.
- For, $AC$ and $CB$ are commensurable in square [Prop. 10.15].
- And twice the (rectangle contained) by $AC$ and $CB$ is commensurable with the (rectangle contained) by $AC$ and $CB$ [Prop. 10.6].
- And thus (the sum of) the squares on $AC$ and $CB$ is incommensurable with twice the (rectangle contained) by $AC$ and $CB$ [Prop. 10.13].
- But, $EG$ is equal to (the sum of) the (squares) on $AC$ and $CB$, and $HK$ equal to twice the (rectangle contained) by $AC$ and $CB$.
- Thus, $EG$ is incommensurable with $HK$.
- Hence, $EH$ is also incommensurable in length with $HN$ [Prop. 6.1], [Prop. 10.11].
- And (they are) rational (straight lines).
- Thus, $EH$ and $HN$ are rational (straight lines which are) commensurable in square only.
- And if two rational (straight lines which are) commensurable in square only are added together then the whole (straight line) is that irrational called binomial [Prop. 10.36].
- Thus, $EN$ is a binomial (straight line) which has been divided (into its component terms) at $H$.
- So, according to the same (reasoning), $EM$ and $MN$ can be shown (to be) rational (straight lines which are) commensurable in square only.
- And $EN$ will (thus) be a binomial (straight line) which has been divided (into its component terms) at the different (points) $H$ and $M$ (which is absurd [Prop. 10.42]).
- And $EH$ is not the same as $MN$, since (the sum of) the (squares) on $AC$ and $CB$ is greater than (the sum of) the (squares) on $AD$ and $DB$.
- But, (the sum of) the (squares) on $AD$ and $DB$ is greater than twice the (rectangle contained) by $AD$ and $DB$ [Prop. 10.59 lem.] .
- Thus, (the sum of) the (squares) on $AC$ and $CB$ - that is to say, $EG$ - is also much greater than twice the (rectangle contained) by $AD$ and $DB$ - that is to say, $MK$.
- Hence, $EH$ is also greater than $MN$ [Prop. 6.1].
- Thus, $EH$ is not the same as $MN$.
- (Which is) the very thing it was required to show.∎

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