Proof: By Euclid

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 show¹.

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It has been shown that the congruence of inscribed angles $\angle{BAC}=\angle{EDF}$ or of central angles $\angle{BGC}=\angle{EHF}$ leads to the congruence of congruent arcs $BKC=ELF$ (editor's note). ↩