Proof: By Euclid
(related to Proposition: 3.26: Equal Angles and Arcs in Equal Circles)
 For let $BC$ and $EF$ have been joined.
 And since circles $ABC$ and $DEF$ are equal, their radii are equal.
 So the two (straight lines) $BG$, $GC$ (are) equal to the two (straight lines) $EH$, $HF$ (respectively).
 And the angle at $G$ (is) equal to the angle at $H$.
 Thus, the base $BC$ is equal to the base $EF$ [Prop. 1.4].
 And since the angle at $A$ is equal to the (angle) at $D$, the segment $BAC$ is thus similar to the segment $EDF$ [Def. 3.11] .
 And they are on equal straight lines [$BC$ and $EF$].
 And similar segments of circles on equal straight lines are equal to one another [Prop. 3.24].
 Thus, segment $BAC$ is equal to (segment) $EDF$.
 And the whole circle $ABC$ is also equal to the whole circle $DEF$.
 Thus, the remaining circumference $BKC$ is equal to the (remaining) circumference $ELF$.
 Thus, in equal circles, equal angles stand upon equal circumferences, whether they are standing at the center or at the circumference.
 (Which is) the very thing which it was required to show^{1}.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes