Proof: By Euclid

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.

fig044e

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.
∎

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References

Adapted from (subject to copyright, with kind permission)

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

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)