There exists one number in mean proportion to two (given)^{1} square numbers. And (one) square (number) has to the (other) square (number) a squared^{2} ratio with respect to (that) the side (of the former has) to the side (of the latter). * Let $A$ and $B$ be square numbers, and let $C$ be the side of $A$, and $D$ (the side) of $B$. * I say that there exists one number in mean proportion to $A$ and $B$, and that $A$ has to $B$ a squared ratio with respect to (that) $C$ (has) to $D$. * So I say that $A$ also has to $B$ a squared ratio with respect to (that) $C$ (has) to $D$.
(not yet contributed)
Proofs: 1
In other words, given two square numbers, their geometric mean is a number (i.e. a positive integer). ↩
Literally, "double" (translator's note). ↩