And let $NO$ have been drawn through $N$, parallel to $CD$.
Thus, $FO$ and $LN$ are each equal to the (rectanglecontained) by $AG$ and $GB$.
And since the (sum of the squares) on $AG$ and $GB$ is rational, and $DM$ is equal to the (sum of the squares) on $AG$ and $GB$, $DM$ is thus rational.
And it has been applied to the rational (straight line) $CD$, producing $CM$ as breadth.