# Proof: By Euclid

• 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.

Thank you to the contributors under CC BY-SA 4.0!

Github:

#### Footnotes

1. The Greek text has "$CD$, $BC$", which is obviously a mistake (translator's note).

2. The Greek text has "$ABCD$", which is obviously a mistake (translator's note).