Parallelograms which are on equal bases and in the same parallels are equal (in area) to one another.

Let $ABCD$, $EFGH$ be parallelograms which are on equal bases $BC$, $FG$ and in the same parallels $AH$, $BG$;I say that the parallelogram $ABCD$ is equal to $EFGH$.For let $BE$, $CH$ be joined.Then, since $BC$ is equal to $FG$while $FG$ is equal to $EH$,$BC$ is also equal to $EH$. [C.N. 1]But they are also parallel.And $EB$, $HC$ join them;but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [Prop. 1.33]Therefore $EBCH$ is a parallelogram. [Prop. 1.34]And it is equal to $ABCD$;for it has the same base $BC$ with it, and is in the same parallels $BC$, $AH$ with it. [Prop. 1.35]For the same reason also $EFGH$ is equal to the same $EBCH$;so that the parallelogram $ABCD$ is also equal to $EFGH$. [C.N. 1]Therefore etc.Q.E.D.