Proof: By Euclid

For since (in the first case) $EGB$ is equal to $GHD$, but $EGB$ is equal to $AGH$ [Prop. 1.15], $AGH$ is thus also equal to $GHD$.
And they are alternate (angles).
Thus, $AB$ is parallel to $CD$ [Prop. 1.27].
Again, since (in the second case, the sum of) $BGH$ and $GHD$ is equal to two right angles, and (the sum of) $AGH$ and $BGH$ is also equal to two right angles [Prop. 1.13], (the sum of) $AGH$ and $BGH$ is thus equal to (the sum of) $BGH$ and $GHD$.
Let $BGH$ have been subtracted from both.
Thus, the remainder $AGH$ is equal to the remainder $GHD$.
And they are alternate (angles).
Thus, $AB$ is parallel to $CD$ [Prop. 1.27].
Thus, if a straight line falling across two straight lines makes the external angle equal to the internal and opposite angle on the same side, or (makes) the (sum of the) internal (angles) on the same side equal to two right angles, then the (two) straight lines will be parallel (to one another).
(Which is) the very thing it was required to show.
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