Parallelograms which are on the same base and between the same parallels are equal^{1} to one another. * Let $ABCD$ and $EBCF$ be parallelograms on the same base $BC$, and between the same parallels $AF$ and $BC$. * I say that $ABCD$ is equal to parallelogram $EBCF$.
Parallelograms on the same base (\(\overline{BC}\)) and between the same parallels (\(\overline{AF}\), \(\overline{BC}\)) are equal in area.
Proofs: 1
Here, for the first time, "equal" means "equal in area", rather than "congruent" (translator's note). ↩