In equal circles, angles standing upon equal circumferences are equal to one another, whether they are standing at the center or at the circumference. * For let the angles $BGC$ and $EHF$ at the centers $G$ and $H$, and the (angles) $BAC$ and $EDF$ at the circumferences, stand upon the equal circumferences $BC$ and $EF$, in the equal circles $ABC$ and $DEF$ (respectively). * I say that angle $BGC$ is equal to (angle) $EHF$, and $BAC$ is equal to $EDF$.
Let two given circles be congruent and let some of its arcs be also congruent ($BC=EF$). Then the corresponding inscribed angles are also congruent ($\angle{BAC}=\angle{EDF}$) and the corresponding central angles are also congruent ($\angle{BGC}=\angle{EHF}$) are also congruent.
This is the converse of the Prop. 3.26.
Proofs: 1
Proofs: 1 2 3 4 5 6 7
Propositions: 8
