Proof: By Euclid
(related to Proposition: 1.33: Parallel Equal Segments Determine a Parallelogram)
 Let $BC$ have been joined.
 And since $AB$ is parallel to $CD$, and $BC$ has fallen across them, the alternate angles $ABC$ and $BCD$ are equal to one another [Prop. 1.29].
 And 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} And the angle $ABC$ is equal to the angle $BCD$.
 Thus, the base $AC$ is equal to the base $BD$, and triangle $ABC$ is equal to triangle $DCB$^{2}, and the remaining angles will be equal to the corresponding remaining angles subtended by the equal sides [Prop. 1.4].
 Thus, angle $ACB$ is equal to $CBD$.
 Also, since the straight line $BC$, (in) falling across the two straight lines $AC$ and $BD$, has made the alternate angles ($ACB$ and $CBD$) equal to one another, $AC$ is thus parallel to $BD$ [Prop. 1.27].
 And ($AC$) was also shown (to be) equal to ($BD$).
 Thus, straight lines joining equal and parallel (straight lines) on the same sides are themselves also equal and parallel.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes