Proof: By Euclid
(related to Proposition: 1.34: Opposite Sides and Opposite Angles of Parallelograms)
 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) $DC$, $CB$^{1}, 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 $ACDB$^{2} in half.
 (Which is) the very thing it was required to show.
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Footnotes