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) $DC$, $CB$, 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.
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Footnotes
