Proof: By Euclid
(related to Proposition: 1.15: Opposite Angles on Intersecting Straight Lines)
 For since the straight line $AE$ stands on the straight line $CD$, making the angles $CEA$ and $AED$, the (sum of the) angles $CEA$ and $AED$ is thus equal to two right angles [Prop. 1.13].
 Again, since the straight line $DE$ stands on the straight line $AB$, making the angles $AED$ and $DEB$, the (sum of the) angles $AED$ and $DEB$ is thus equal to two right angles [Prop. 1.13].
 But (the sum of) $CEA$ and $AED$ was also shown (to be) equal to two right angles.
 Thus, (the sum of) $CEA$ and $AED$ is equal to (the sum of) $AED$ and $DEB$ [C.N. 1] .
 Let $AED$ have been subtracted from both.
 Thus, the remainder $CEA$ is equal to the remainder $BED$ [C.N. 3] .
 Similarly, it can be shown that $CEB$ and $DEA$ are also equal.
 Thus, if two straight lines cut one another then they make the vertically opposite angles equal to one another.
 (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"