Proof: By Euclid
(related to Proposition: 7.35: Least Common Multiple Divides Common Multiple)
- For if $E$ does not measure $CD$ then let $E$ leave $CF$ less than itself (in) measuring $DF$.
- And since $A$ and $B$ (both) measure $E$, and $E$ measures $DF$, $A$ and $B$ will thus also measure $DF$.
- And ($A$ and $B$) also measure the whole of $CD$.
- Thus, they will also measure the remainder $CF$, which is less than $E$.
- The very thing is impossible.
- Thus, $E$ cannot not measure $CD$.
- Thus, ($E$) measures ($CD$).
- (Which is) the very thing it was required to show.
Thank you to the contributors under CC BY-SA 4.0!
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"