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

Let $ABCD$, $EBCF$ be parallelograms on the same base $BC$ and in the same parallels $AF$, $BC$;I say that $ABCD$ is equal (in area) to the parallelogram $EBCF$.For, since $ABCD$ is a parallelogram,$AD$ is equal to $BC$. [Prop. 1.34]For the same reason also$EF$ is equal to $BC$,so that $AD$ is also equal to $EF$; [C.N. 1]and $DE$ is common;therefore the whole $AE$ is equal to the whole $DF$. [C.N. 2]But $AB$ is also equal to $DC$; [Prop. 1.34]therefore the two sides $EA$, $AB$ are equal to the two sides $FD$, $DC$ respectively,and the angle $FDC$ is equal to the angle $EAB$, the exterior to the interior; [Prop. 1.29]therefore the base $EB$ is equal to the base $FC$,and the triangle $EAB$ will be equal to the triangle $FDC$. [Prop. 1.4]Let $DGE$ be subtracted from each;therefore the trapezium $ABGD$ which remains is equal to the trapezium $EGCF$ which remains. [C.N. 3]Let the triangle $GBC$ be added to each;therefore the whole parallelogram $ABCD$ is equal to the whole parallelogram $EBCF$. [C.N. 2]Therefore etc.Q.E.D.