Proof: By Euclid
(related to Proposition: 1.40: Triangles of Equal Area IV)
- For let $AD$ have been joined.
- I say that $AD$ is parallel to $BE$.
- For if not, let $AF$ have been drawn through $A$ parallel to $BE$ [Prop. 1.31], and let $FE$ have been joined.
- Thus, triangle $ABC$ is equal to triangle $FCE$.
- For they are on equal bases, $BC$ and $CE$, and between the same parallels, $BE$ and $AF$ [Prop. 1.38].
- But, triangle $ABC$ is equal to [triangle] $DCE$.
- Thus, [triangle] $DCE$ is also equal to triangle $FCE$, the greater to the lesser.
- The very thing is impossible.
- Thus, $AF$ is not parallel to $BE$.
- Similarly, we can show that neither (is) any other (straight line) than $AD$.
- Thus, $AD$ is parallel to $BE$.
- Thus, equal triangles which are on equal bases, and on the same side, are also between the same parallels.
- (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"