# Proof: By Euclid

(related to Proposition: 1.37: Triangles of Equal Area I)

• Let $AD$ have been produced in both directions to $E$ and $F$, and let the (straight line) $BE$ have been drawn through $B$ parallel to $CA$ [Prop. 1.31], and let the (straight line) $CF$ have been drawn through $C$ parallel to $BD$ [Prop. 1.31].
• Thus, $EBCA$ and $DBCF$ are both parallelograms, and are equal.
• For they are on the same base $BC$, and between the same parallels $BC$ and $EF$ [Prop. 1.35].
• And the triangle $ABC$ is half of the parallelogram $EBCA$.
• For the diagonal $AB$ cuts the latter in half [Prop. 1.34].
• And the triangle $DBC$ (is) half of the parallelogram $DBCF$.
• For the diagonal $DC$ cuts the latter in half [Prop. 1.34].
• And the halves of equal things are equal to one another.]1 Thus, triangle $ABC$ is equal to triangle $DBC$.
• Thus, triangles which are on the same base and between the same parallels are equal to one another.
• (Which is) the very thing it was required to show.

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:
@Fitzpatrick

### References

#### Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"

#### Footnotes

1. This is an additional common notion (translator's note).