If a parallelogram has the same base as a triangle, and is between the same parallels, then the parallelogram is double (the area) of the triangle.  For let parallelogram $ABCD$ have the same base $BC$ as triangle $EBC$, and let it be between the same parallels, $BC$ and $AE$. I say that  parallelogram $ABCD$ is double (the area) of triangle $BEC$.  For let $AC$ have been joined. So triangle $ABC$ is equal to triangle $EBC$. For it is on the same base, $BC$,  as ($EBC$), and between the same parallels, $BC$ and $AE$ [Prop.~1.37]. But, parallelogram $ABCD$ is double (the area) of triangle $ABC$. For the diagonal $AC$ cuts the former in half [Prop.~1.34]. So parallelogram $ABCD$ is also double (the area) of triangle $EBC$.  Thus, if a parallelogram has the same base as a triangle, and is between the same parallels, then the parallelogram is double (the area) of the triangle. (Which is) the very thing it was required to show.