In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

Let $ACDB$ be a parallelogrammic area, and $BC$ its diameter.I say that the opposite sides and angles of the parallelogram $ACDB$ are equal to one another, and the diameter $BC$ bisects it.For, since $AB$ is parallel to $CD$, and the straight line $BC$ has falled upon them,the alternate angles $ABC$, $BCD$ are equal to one another. [Prop. 1.29]Again, since $AC$ is parallel to $BD$, and $BC$ has fallen upon them,the alternate angles $ACB$, $CBD$ are equal to one another. [Prop. 1.29]Therefore $ABC$, $DCB$ are two triangles having the two angles $ABC$, $BCA$ euqal to the two angles $DCB$, $CBD$ respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, $BC$;therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [Prop. 1.26]therefore the side $AB$ is equal to $CD$,and $AC$ to $BD$,and further the angle $BAC$ is equal to the angle $CDB$.And since the angle $ABC$ is equal to the angle $BCD$,and the angle $CBD$ is equal to the angle $ACB$,the whole angle $ABD$ is equal to the whole angle $ACD$. [C.N. 2]And the angle $BAC$ was also proved equal to the angle $CDB$.Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.I say, next, that the diameter also bisects the areas.For, since $AB$ is equal to $CD$,and $BC$ is common,the two sides $AB$, $BC$ are equal to the two sides $DC$, $CB$ respectively;and the angle $ABC$ is equal to the angle $BCD$;therefore the base $AC$ is also equal to $DB$,and the triangle $ABC$ is equal to the triangle $DCB$; [Prop. 1.4]Therefore the diameter $BC$ bisects the parallelogram $ACDB$.Q.E.D.