Triangles which are on equal bases and in the same parallels are equal to one another.

Let $ABC$, $DEF$ be triangles on equal bases $BC$, $EF$ and in the same parallels $BF$, $AD$;I say that the triangle $ABC$ is equal to the triangle $DEF$.For let $AD$ be produced in both directions to $G$, $H$;through $B$ let $BG$ be drawn parallel to $CA$, [Prop. 1.31]and through $F$ let $FH$ be drawn parallel to $DE$.Then each of the figures $GBCA$, $DEFH$ is a parallelogram;and $GBCA$ is equal to $DEFH$;for they are on equal bases $BC$, $EF$ and in the same parallels $BF$, $GH$. [Prop. 1.36]Moreover the triangle $ABC$ is half of the parallelogram $GBCA$; for the diameter $AB$ bisects it. [Prop. 1.34]And the triangle $FED$ is half of the paralleogram $DEFH$; for the diameter $DF$ bisects it. [Prop. 1.34][But the halves of equal things are equal to one another.]Therefore the triangle $ABC$ is equal to the triangle $DEF$.Therefore etc.Q.E.D.