If there are any multitude whatsoever of numbers in continued proportion, and each makes some (number by) multiplying itself, then the (numbers) created from them will (also) be in (continued) proportion. And if the original (numbers) make some (more numbers by) multiplying the created (numbers) then these will also be in (continued) proportion [and this always happens with the extremes]. * Let $A$, $B$, $C$ be any multitude whatsoever of numbers in continued proportion, (such that) as $A$ (is) to $B$, so $B$ (is) to $C$. * And let $A$, $B$, $C$ make $D$, $E$, $F$ (by) multiplying themselves, and let them make $G$, $H$, $K$ (by) multiplying $D$, $E$, $F$. * I say that $D$, $E$, $F$ and $G$, $H$, $K$ are in continued proportion.
(not yet contributed)
Proofs: 1