In equal circles, equal angles stand upon equal circumferences whether they are standing at the center or at the circumference. * Let $ABC$ and $DEF$ be equal circles, and within them let $BGC$ and $EHF$ be equal angles at the center, and $BAC$ and $EDF$ (equal angles) at the circumference. * I say that circumference $BKC$ is equal to circumference $ELF$.
Let two given circles be congruent and let some of its inscribed angles be congruent ($\angle{BAC}=\angle{EDF}$) or some of its central angles be congruent ($\angle{BGC}=\angle{EHF}$) be congruent. Then the corresponding arcs are also congruent ($BKC=ELF$).
Proofs: 1
Proofs: 1 2 3 4 5
Propositions: 6
