Proof: By Euclid
(related to Proposition: 1.13: Angles at Intersections of Straight Lines)
 In fact, if $CBA$ is equal to $ABD$ then they are two right angles [Def. 1.10] .
 But, if not, let $BE$ have been drawn from the point $B$ at right angles to [the straight line $CD$ [Prop. 1.11].
 Thus, $CBE$ and $EBD$ are two right angles.
 And since $CBE$ is equal to the two (angles) $CBA$ and $ABE$, let $EBD$ have been added to both.
 Thus, the (sum of the angles) $CBE$ and $EBD$ is equal to the (sum of the) three (angles) $CBA$, $ABE$, and $EBD$ [C.N. 2] .
 Again, since $DBA$ is equal to the two (angles) $DBE$ and $EBA$, let $ABC$ have been added to both.
 Thus, the (sum of the angles) $DBA$ and $ABC$ is equal to the (sum of the) three (angles) $DBE$, $EBA$, and $ABC$ [C.N. 2] .
 But (the sum of) $CBE$ and $EBD$ was also shown (to be) equal to the (sum of the) same three (angles).
 And things equal to the same thing are also equal to one another [C.N. 1] .
 Therefore, (the sum of) $CBE$ and $EBD$ is also equal to (the sum of) $DBA$ and $ABC$.
 But, (the sum of) $CBE$ and $EBD$ is two right angles.
 Thus, (the sum of) $ABD$ and $ABC$ is also equal to two right angles.
 Thus, if a straight line stood on a(nother) straight line makes angles, it will certainly either make two right angles, or (angles whose sum is) equal to two right angles.
 (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"