Parallelograms which are on the same base and between the same parallels are equal 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 $EBCF$.  For since $ABCD$ is a parallelogram, $AD$ is equal to $BC$ [Prop.~1.34]. So, for the same (reasons), $EF$ is also equal to $BC$. So $AD$ is also equal to $EF$. And $DE$ is common. Thus, the whole (straight-line) $AE$ is equal to the whole (straight-line) $DF$.  And $AB$ is also equal to $DC$. So the two (straight-lines) $EA$, $AB$ are equal to the two (straight-lines) $FD$, $DC$, respectively. And angle $FDC$ is equal to angle $EAB$, the external to the internal [Prop.~1.29]. Thus, the base $EB$ is equal to the base $FC$, and triangle $EAB$ will be equal to triangle $DFC$ [Prop.~1.4]. Let $DGE$ have been taken away from both.  Thus, the remaining trapezium $ABGD$ is equal to the remaining trapezium $EGCF$. Let triangle $GBC$ have been added to both. Thus, the whole parallelogram $ABCD$ is equal to the whole parallelogram $EBCF$.  Thus, parallelograms 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.