Triangles which are on the same base and between the same parallels are equal to one another.    Let $ABC$ and $DBC$ be triangles on the same base $BC$, and between the same parallels $AD$ and $BC$. I say that triangle $ABC$ is equal to triangle $DBC$.  Let $AD$ have been produced in both directions to $E$ and $F$, and let the (straight-line) $BE$ have been drawn through $B$ parallel to $CA$ [Prop.~1.31], and let the (straight-line) $CF$ have been drawn through $C$ parallel to $BD$ [Prop.~1.31]. Thus, $EBCA$ and $DBCF$ are both parallelograms, and are equal. For they are on the same base $BC$, and between the same parallels $BC$ and $EF$ [Prop.~1.35]. And the triangle $ABC$ is half of  the parallelogram $EBCA$. For the diagonal $AB$ cuts the latter in half [Prop.~1.34]. And the  triangle $DBC$ (is) half of the parallelogram $DBCF$. For the diagonal $DC$ cuts the latter in half [Prop.~1.34]. [And the halves of equal things are equal to one another.] Thus, triangle $ABC$ is equal to triangle $DBC$.  Thus, triangles which are on the same base and between the same parallels are equal to one another. (Which is) the very thing it was required to show.