Triangles which are on equal bases and between the same parallels are equal to one another. Let $ABC$ and $DEF$ be triangles on the equal bases $BC$ and $EF$, and between the same parallels $BF$ and $AD$. I say that triangle $ABC$ is equal to triangle $DEF$.  For let $AD$ have been produced in both directions to $G$ and $H$, and let the (straight-line) $BG$ have been drawn through $B$ parallel to $CA$ [Prop.~1.31], and let the (straight-line) $FH$ have been drawn through $F$ parallel to $DE$ [Prop.~1.31]. Thus,  $GBCA$ and $DEFH$ are each parallelograms. And $GBCA$ is equal to $DEFH$. For they are on the equal bases $BC$ and $EF$, and between  the same parallels $BF$ and $GH$ [Prop.~1.36]. And triangle $ABC$ is half of the parallelogram $GBCA$. For the diagonal $AB$ cuts the latter in half [Prop.~1.34]. And triangle $FED$ (is) half of parallelogram $DEFH$. For the diagonal $DF$ cuts the latter in half. [And the halves of equal things are equal to one another.] Thus, triangle $ABC$ is equal to triangle $DEF$.  Thus, triangles 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.