For, $AB$ and $CD$ being two unequal magnitudes, and $AB$ (being) the lesser, let the remainder never measure the (magnitude) before it, (when) the lesser (magnitude is) continually subtracted in turn from the greater.
For if they are commensurable then some magnitude will measure them (both).
If possible, let it (so) measure (them), and let it be $E$.
And let $AB$ leave $CF$ less than itself (in) measuring $FD$, and let $CF$ leave $AG$ less than itself (in) measuring $BG$, and let this happen continually, until some magnitude which is less than $E$ is left.
Let (this) have occurred,1 and let $AG$, (which is) less than $E$, have been left.
Therefore, since $E$ measures $AB$, but $AB$ measures $DF$, $E$ will thus also measure $FD$.
And it also measures the whole (of) $CD$.
Thus, it will also measure the remainder $CF$.
But, $CF$ measures $BG$.
Thus, $E$ also measures $BG$.
And it also measures the whole (of) $AB$.
Thus, it will also measure the remainder $AG$, the greater (measuring) the lesser.