Proof: By Euclid
(related to Proposition: 6.14: Characterization of Congruent Parallelograms)
 For let the parallelogram $FE$ have been completed.
 Therefore, since parallelogram $AB$ is equal to parallelogram $BC$, and $FE$ (is) some other (parallelogram), thus as (parallelogram) $AB$ is to $FE$, so (parallelogram) $BC$ (is) to $FE$ [Prop. 5.7].
 But, as (parallelogram) $AB$ (is) to $FE$, so $DB$ (is) to $BE$, and as (parallelogram) $BC$ (is) to $FE$, so $GB$ (is) to $BF$ [Prop. 6.1].
 Thus, also, as $DB$ (is) to $BE$, so $GB$ (is) to $BF$.
 Thus, in parallelograms $AB$ and $BC$ the sides about the equal angles are reciprocally proportional.
 And so, let $DB$ be to $BE$, as $GB$ (is) to $BF$.
 I say that parallelogram $AB$ is equal to parallelogram $BC$.
 For since as $DB$ is to $BE$, so $GB$ (is) to $BF$, but as $DB$ (is) to $BE$, so parallelogram $AB$ (is) to parallelogram $FE$, and as $GB$ (is) to $BF$, so parallelogram $BC$ (is) to parallelogram $FE$ [Prop. 6.1], thus, also, as (parallelogram) $AB$ (is) to $FE$, so (parallelogram) $BC$ (is) to $FE$ [Prop. 5.11].
 Thus, parallelogram $AB$ is equal to parallelogram $BC$ [Prop. 5.9].
 Thus, in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional.
 And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
 (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"