In parallelogrammic figures the opposite sides and angles are equal to one another, and a diagonal cuts them in half.  Let $ACDB$ be a parallelogrammic figure, and $BC$ its diagonal. I say that for parallelogram $ACDB$, the opposite sides and angles are equal to one another, and the diagonal $BC$ cuts it in half.  For since $AB$ is parallel to $CD$, and the straight-line $BC$ has fallen across  them, the alternate angles $ABC$ and $BCD$ are equal to one another [Prop.~1.29].  Again, since $AC$ is parallel to $BD$, and $BC$ has fallen across them, the alternate angles $ACB$ and $CBD$ are equal to one another [Prop.~1.29].  So $ABC$ and $BCD$ are two triangles having the two angles $ABC$ and $BCA$ equal to the two (angles) $BCD$ and $CBD$, respectively, and one side equal to one side---the (one) by the equal angles and common to them, (namely) $BC$. Thus, they will also  have the remaining sides  equal to the corresponding remaining (sides), and the remaining angle (equal) to the remaining angle [Prop.~1.26]. Thus, side $AB$ is equal to $CD$, and $AC$ to $BD$. Furthermore, angle $BAC$ is  equal to $CDB$. And since angle $ABC$ is equal to $BCD$, and $CBD$ to $ACB$, the whole (angle) $ABD$ is thus equal to the whole (angle) $ACD$. And  $BAC$ was also shown  (to be) equal to $CDB$.  Thus, in parallelogrammic figures the opposite sides and angles are equal to one another.  And, I also say that a diagonal cuts them in half. For since $AB$ is equal to $CD$, and $BC$ (is) common, the two (straight-lines) $AB$, $BC$ are equal to the two (straight-lines) $DC$, $CB$, respectively. And angle $ABC$ is equal to angle $BCD$. Thus, the base $AC$ (is) also equal to $DB$, and triangle $ABC$ is equal to triangle $BCD$ [Prop.~1.4].  Thus, the diagonal $BC$ cuts the parallelogram $ACDB$ in half. (Which is) the very thing it was required to show.