Proof: By Euclid

For if $BGC$ is unequal to $EHF$, one of them is greater.
Let $BGC$ be greater, and let the (angle) $BGK$, equal to angle $EHF$, have been constructed on the straight line $BG$, at the point $G$ on it [Prop. 1.23].
But equal angles (in equal circles) stand upon equal circumferences, when they are at the centers [Prop. 3.26].
Thus, circumference $BK$ (is) equal to circumference $EF$.
But, $EF$ is equal to $BC$.
Thus, $BK$ is also equal to $BC$, the lesser to the greater.
The very thing is impossible.
Thus, angle $BGC$ is not unequal to $EHF$.
Thus, (it is) equal.
And the (angle) at $A$ is half $BGC$, and the (angle) at $D$ half $EHF$ [Prop. 3.20].
Thus, the angle at $A$ (is) also equal to the (angle) at $D$.
Thus, in equal circles, angles standing upon equal circumferences are equal to one another, whether they are standing at the center or at the circumference.
(Which is) the very thing it was required to show.¹

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It has been shown that in congruent circles the equality of arcs ($BC=EF$) leads to the congruence of the corresponding inscribed angles ($\angle{BAC}=\angle{EDF}$) and the congruence of the corresponding central angles ($\angle{BGC}=\angle{EHF}$) (editor's note). ↩