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

Let $ABC$, $DBC$ be triangles on the same base $BC$ and in the same parallels $AD$, $BC$;I say that the triangle $ABC$ is euqal to the triangle $DBC$.Let $AD$ be produced in both directions to $E$, $F$;through $B$ let $BE$ be drawn parallel to $CA$, [Prop. 1.31]and through $C$ let $CF$ be drawn parallel to $BD$. [Prop. 1.31]Then each of the figures $EBCA$, $DBCF$ is a parallelogram;and they are equal,for they are on the same base $BC$ and in the same parallels $BC$, $EF$. [Prop. 1.35]Moreover the triangle $ABC$ is half of the parallelogram $EBCA$; for the diameter $AB$ bisects it. [Prop. 1.34]And the triangle $DBC$ is half of the parallelogram $DBCF$; for the diameter $DC$ 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 $DBC$.Therefore etc.Q.E.D.