In equal circle,arcs which subtend eques, either at the centres or at the circumferences, are equal.
Let ABE, DEF be equal circles, adn let the BGC=the EHF at the centres ; and consequently
the BAC= the EDF at the circles.
It is required to prove that the Arc BKC= thearc ELF.
Proof. Apply the ABC to the DEF, so that the centre G falls on the centre H, and GB falls along HE.
Then because the BGC= the EHF,
GC will fall along HF.
And because the circles have equal radii, B will fall on E, and C on F, and the circumferences of the circles will coincide entirely.
the arc BKC must coincide with the arc ELF ;
the arc BKC=the arc ELF.
Corollary. In equal circles sectors which have equal angles are equal.
Obs. It is clear that any theorem relating to arcs, angles, and chords in equal circles must also be true in the same circle.