If three straight lines are in (continued) proportion then the parallelepipedal solid (formed) from the three (straight lines) is equal to the equilateral parallelepipedal solid on the middle (straight line which is) equiangular to the aforementioned (parallelepipedal solid). * Let $A$, $B$, and $C$ be three straight lines in (continued) proportion, (such that) as $A$ (is) to $B$, so $B$ (is) to $C$. * I say that the (parallelepiped) solid (formed) from $A$, $B$, and $C$ is equal to the equilateral solid on $B$ (which is) equiangular with the aforementioned (solid).
(not yet contributed)
Proofs: 1