Let $AB$ and $BC$ be two squares, and let them be laid down such that $DB$ is straight-on to $BE$. $FB$ is, thus, also straight-on to $BG$. And let the parallelogram $AC$ have been completed. I say that $AC$ is a square, and that $DG$ is in mean proportion to $AB$ and $BC$, and, moreover, $DC$ is in mean proportion to $AC$ and $CB$.
(not yet contributed)