Parallelograms which are on equal bases and between the same parallels are equal to one another. Let $ABCD$ and $EFGH$ be parallelograms which are on the equal bases $BC$ and $FG$, and (are) between the same parallels $AH$ and $BG$. I say that the parallelogram $ABCD$ is equal to $EFGH$.  For let $BE$ and $CH$ have been joined. And since $BC$ is equal to $FG$, but $FG$ is equal to $EH$ [Prop.~1.34], $BC$ is thus equal to $EH$. And they are also parallel, and $EB$ and $HC$ join them. But (straight-lines) joining equal and parallel (straight-lines) on the same sides are (themselves) equal and parallel [Prop.~1.33] [thus, $EB$ and $HC$ are also equal and parallel]. Thus, $EBCH$ is a parallelogram [Prop.~1.34], and is equal to $ABCD$. For it has  the same base, $BC$, as ($ABCD$), and is between the same parallels, $BC$ and $AH$, as ($ABCD$) [Prop.~1.35]. So, for the same (reasons), $EFGH$ is also equal to the same (parallelogram) $EBCH$ [Prop.~1.34].  So that the parallelogram $ABCD$ is  also equal to $EFGH$.  Thus, parallelograms which are on equal bases and between the same parallels are equal to one another. (Which is) the very thing it was required to show.