Proof: By Euclid

For since $AB$ is greater than $C$, let $BE$ be made equal to $C$.
So, the lesser of $AE$ and $EB$, being multiplied, will sometimes be greater than $D$ [Def. 5.4] .
First of all, let $AE$ be less than $EB$, and let $AE$ have been multiplied, and let $FG$ be a multiple of it which (is) greater than $D$.
And as many times as $FG$ is (divisible) by $AE$, so many times let $GH$ also have become (divisible) by $EB$, and $K$ by $C$.
And let the double multiple $L$ of $D$ have been taken, and the triple multiple $M$, and several more, (each increasing) in order by one, until the (multiple) taken becomes the first multiple of $D$ (which is) greater than $K$.
Let it have been taken, and let it also be the quadruple multiple $N$ of $D$ - the first (multiple) greater than $K$.
Therefore, since $K$ is less than $N$ first, $K$ is thus not less than $M$.
And since $FG$ and $GH$ are equal multiples of $AE$ and $EB$ (respectively), $FG$ and $FH$ are thus equal multiples of $AE$ and $AB$ (respectively) [Prop. 5.1].
And $FG$ and $K$ are equal multiples of $AE$ and $C$ (respectively).
Thus, $FH$ and $K$ are equal multiples of $AB$ and $C$ (respectively).
Thus, $FH$, $K$ are equal multiples of $AB$, $C$.
Again, since $GH$ and $K$ are equal multiples of $EB$ and $C$, and $EB$ (is) equal to $C$, $GH$ (is) thus also equal to $K$.
And $K$ is not less than $M$.
Thus, $GH$ not less than $M$ either.
And $FG$ (is) greater than $D$.
Thus, the whole of $FH$ is greater than $D$ and $M$ (added) together.
But, $D$ and $M$ (added) together is equal to $N$, inasmuch as $M$ is three times $D$, and $M$ and $D$ (added) together is four times $D$, and $N$ is also four times $D$.
Thus, $M$ and $D$ (added) together is equal to $N$.
But, $FH$ is greater than $M$ and $D$.
Thus, $FH$ exceeds $N$.
And $K$ does not exceed $N$.
And $FH$, $K$ are equal multiples of $AB$, $C$, and $N$ another random multiple of $D$.
Thus, $AB$ has a greater ratio to $D$ than $C$ (has) to $D$ [Def. 5.7] .
So, I say that $D$ also has a greater ratio to $C$ than $D$ (has) to $AB$.
For, similarly, by the same construction, we can show that $N$ exceeds $K$, and $N$ does not exceed $FH$.
And $N$ is a multiple of $D$, and $FH$, $K$ other random equal multiples of $AB$, $C$ (respectively).
Thus, $D$ has a greater ratio to $C$ than $D$ (has) to $AB$ [Def. 5.5] .
And so let $AE$ be greater than $EB$.
So, the lesser, $EB$, being multiplied, will sometimes be greater than $D$.
Let it have been multiplied, and let $GH$ be a multiple of $EB$ (which is) greater than $D$.
And as many times as $GH$ is (divisible) by $EB$, so many times let $FG$ also have become (divisible) by $AE$, and $K$ by $C$.
So, similarly (to the above), we can show that $FH$ and $K$ are equal multiples of $AB$ and $C$ (respectively).
And, similarly (to the above), let the multiple $N$ of $D$, (which is) the first (multiple) greater than $FG$, have been taken.
So, $FG$ is again not less than $M$.
And $GH$ (is) greater than $D$.
Thus, the whole of $FH$ exceeds $D$ and $M$, that is to say $N$.
And $K$ does not exceed $N$, inasmuch as $FG$, which (is) greater than $GH$ - that is to say, $K$ - also does not exceed $N$.
And, following the above (arguments), we (can) complete the proof in the same manner.
Thus, for unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude).
And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater.
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!

Github:
non-Github:: @Fitzpatrick

References

Adapted from (subject to copyright, with kind permission)

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

◀ ▲ ▶Branches / Geometry / Elements-euclid / Book--5-proportion / Proof: By Euclid

Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)