If there are any number of magnitudes whatsoever, and (some) other (magnitudes) of equal number 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 number of magnitudes whatsoever, $A$, $B$, $C$, and (some) other (magnitudes), $D$, $E$, $F$, of equal number to them, (which are) in the same ratio taken two by two, (so 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 they will also be in the same ratio via equality. * (That is, as $A$ is to $C$, so $D$ is to $F$.)
In modern notation, this proposition reads that if \[\frac\alpha\beta=\frac\epsilon\zeta\text{ and }\frac\beta\gamma=\frac\zeta\eta\text{ and }\frac\gamma\delta=\frac\eta\theta,\] then \[\frac\alpha\delta=\frac\epsilon\theta,\]
for all positive real numbers \(\alpha,\beta,\gamma,\delta,\epsilon,\zeta,\eta,\theta\).
Proofs: 1
Definitions: 1
Proofs: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Sections: 19
