And let $CH$, equal to the (square) on $AG$, have been applied to $CD$, producing $CK$ as breadth, and $KL$, equal to the (square) on $BG$, producing $KM$ as breadth.
Thus, the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$.
And the sum of the (squares) on $AG$ and $GB$ is rational.
And since the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, of which $CE$ is equal to the (square) on $AB$, the remainder $FL$ is thus equal to twice the (rectanglecontained) by $AG$ and $GB$ [Prop. 2.7].
Therefore, let $FM$ have been cut in half at point $N$.
And let $NO$ have been drawn through $N$, parallel to either of $CD$ or $ML$.
Thus, $FO$ and $NL$ are each equal to the (rectanglecontained) by $AG$ and $GB$.