In equal circles angles standing on equal circumferences equal one another whether they stand at the centers or at the circumferences.

For in equal circles *ABC* and *DEF,* on equal circumferences *BC* and *EF,* let the angles *BGC* and *EHF* stand at the centers *G* and *H,* and the angles *BAC* and *EDF* at the circumferences.

I say that the angle *BGC* equals the angle *EHF,* and the angle *BAC* equals the angle *EDF.*

For, if the angle *BGC* does not equal the angle *EHF,* one of them is greater. Let the angle *BGC* be greater. Construct the angle *BGK* equal to the angle *EHF* on the straight line *BG* and at the point *G* on it.

Now equal angles stand on equal circumferences when they are at the centers, therefore the circumference *BK* equals the circumference *EF.*

But *EF* equals *BC,* therefore *BK* also equals *BC,* the less equals the greater, which is impossible.

Therefore the angle *BGC* is not unequal to the angle *EHF,* therefore it equals it.

And the angle at *A* is half of the angle *BGC,* and the angle at *D* half of the angle *EHF,* therefore the angle at *A* also equals the angle at *D.*

Therefore *in equal circles angles standing on equal circumferences equal one another whether they stand at the centers or at the circumferences.
*

Q.E.D.