Proof: By Euclid

(not yet contributed)

fig005e

For if $A$ and $B$ are commensurable (magnitudes) then some magnitude will measure them.
Let it (so) measure (them), and let it be $C$.
And as many times as $C$ measures $A$, so many units let there be in $D$.
And as many times as $C$ measures $B$, so many units let there be in $E$.
Therefore, since $C$ measures $A$ according to the units in $D$, and a unit also measures $D$ according to the units in it, a unit thus measures the number $D$ as many times as the magnitude $C$ (measures) $A$.
Thus, as $C$ is to $A$, so a unit (is) to $D$¹ [Def. 7.20] .
Thus, inversely, as $A$ (is) to $C$, so $D$ (is) to a unit [Prop. 5.7 corr.] .
Again, since $C$ measures $B$ according to the units in $E$, and a unit also measures $E$ according to the units in it, a unit thus measures $E$ the same number of times that $C$ (measures) $B$.
Thus, as $C$ is to $B$, so a unit (is) to $E$ [Def. 7.20] .
And it was also shown that as $A$ (is) to $C$, so $D$ (is) to a unit.
Thus, via equality, as $A$ is to $B$, so the number $D$ (is) to the (number) $E$ [Prop. 5.22].
Thus, the commensurable magnitudes $A$ and $B$ have to one another the ratio which the number $D$ (has) to the number $E$.
(Which is) the very thing it was required to show.

∎

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There is a slight logical gap here, since [Def. 7.20] applies to four numbers, rather than two numbers and two magnitudes (translator's note). ↩