In equal circles, equal straight lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser. * Let $ABC$ and $DEF$ be equal circles, and let $AB$ and $DE$ be equal straight lines in these circles, cutting off the greater circumferences $ACB$ and $DFE$, and the lesser (circumferences) $AGB$ and $DHE$ (respectively). * I say that the greater circumference $ACB$ is equal to the greater circumference $DFE$, and the lesser circumference $AGB$ to (the lesser) $DHE$.
The arcs are congruent (the longer ones ${ACB}\cong {DFE}$ and the shorter ones ${AGB}\cong {DHE}$) if segments with equal lengths ($|\overline{AB} | = | \overline{DE} |$) connect their endpoints in congruent circles.
Proofs: 1