Proof: By Euclid

For let the centers of the circles have been found [Prop. 3.1], and let them be (at) $K$ and $L$.
And let $BK$, $KC$, $EL$, and $LF$ have been joined.
And since the circumference $BGC$ is equal to the circumference $EHF$, the angle $BKC$ is also equal to (angle) $ELF$ [Prop. 3.27].
And since the circles $ABC$ and $DEF$ are equal, their radii are also equal [Def. 3.1] .
So the two (straight lines) $BK$, $KC$ are equal to the two (straight lines) $EL$, $LF$ (respectively).
And they contain equal angles.
Thus, the base $BC$ is equal to the base $EF$ [Prop. 1.4].
Thus, in equal circles, equal straight lines subtend equal circumferences.
(Which is) the very thing it was required to show.¹

∎

Thank you to the contributors under CC BY-SA 4.0!

It has been shown that the segments ($\overline{BC}$, $\overline{EF}$) connecting the endpoints of two congruent arcs in two congruent circles have equal lengths (editor's note). ↩