If there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality. * Let $A$, $B$, and $C$ be three magnitudes, and $D$, $E$ and $F$ other (magnitudes) of equal number to them, (being) in the same ratio taken two by two. * And let their proportion be perturbed, (so that) as $A$ (is) to $B$, so $E$ (is) to $F$, and as $B$ (is) to $C$, so $D$ (is) to $E$. * I say that as $A$ is to $C$, so $D$ (is) to $F$.
In modern notation, this proposition reads that if \[\frac\alpha\beta=\frac\epsilon\zeta\text{ and }\frac\beta\gamma=\frac\delta\epsilon,\] then \[\frac\alpha\gamma=\frac\delta\zeta,\]
for all positive real numbers \(\alpha,\beta,\gamma,\delta,\epsilon,\zeta\).
Proofs: 1
Sections: 1
