If there are any multitude of numbers whatsoever, and (some) other (numbers) of equal multitude to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality. * Let there be any multitude of numbers whatsoever, $A$, $B$, $C$, and (some) other (numbers), $D$, $E$, $F$, of equal multitude to them, (which are) in the same ratio taken two by two, (such that) as $A$ (is) to $B$, so $D$ (is) to $E$, and as $B$ (is) to $C$, so $E$ (is) to $F$. * I say that also, via equality, as $A$ is to $C$, so $D$ (is) to $F$.
In modern notation, this proposition states that if \[\frac AB=\frac DE\wedge \frac BC=\frac EF,\] then \[\frac AC=\frac DF.\]
Proofs: 1