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 $GB$, producing $KM$ as breadth.
Thus, the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$.
And since $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, of which the (square) on $AB$ is equal to $CE$, the remainder, twice the (rectanglecontained) by $AG$ and $GB$, is thus equal to $FL$ [Prop. 2.7].
Therefore, since the (sum of the squares) on $AG$ and $GB$ - that is to say, $CL$ - is medial, and twice the (rectanglecontained) by $AG$ and $GB$ - that is to say, $FL$ - (is) rational, $CL$ is thus incommensurable with $FL$.
And as $CL$ (is) to $FL$, so $CM$ is to $FM$ [Prop. 6.1].