Proof: By Euclid
(related to Proposition: 3.28: Straight Lines Cut Off Equal Arcs in Equal Circles)
 For let the centers of the circles, $K$ and $L$, have been found [Prop. 3.1], and let $AK$, $KB$, $DL$, and $LE$ have been joined.
 And since ($ABC$ and $DEF$) are equal circles, their radii are also equal [Def. 3.1] .
 So the two (straight lines) $AK$, $KB$ are equal to the two (straight lines) $DL$, $LE$ (respectively).
 And the base $AB$ (is) equal to the base $DE$.
 Thus, angle $AKB$ is equal to angle $DLE$ [Prop. 1.8].
 And equal angles stand upon equal circumferences, when they are at the centers [Prop. 3.26].
 Thus, circumference $AGB$ (is) equal to $DHE$.
 And the whole circle $ABC$ is also equal to the whole circle $DEF$.
 Thus, the remaining circumference $ACB$ is also equal to the remaining circumference $DFE$.
 Thus, in equal circles, equal straight lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser.
 (Which is) the very thing it was required to show.^{1}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes