Angles on Equal Arcs are Equal

Theorem

In equal circles, angles standing on equal arcs are equal to one another, whether at the center or at the circumference of those circles.

Proof

Let $ABC$ and $DEF$ be equal circles.

Let $\angle BGC$ and $\angle EHF$ stand on the equal arcs $BC$ and $EF$.

Suppose $\angle BGC \ne \angle EHF$.

Then one of them is bigger. Suppose $\angle BGC > \angle EHF$.

On the straight line $BG$, construct $\angle BGK$ equal to $\angle EHF$.

From Equal Angles in Equal Circles, these angles stand on equal arcs.

So arcs $BK = EF$.

But $EF = BC$ and so $BK = BC$, which is impossible.

So $\angle BGC$ and $\angle EHF$ are not unequal, therefore $\angle BGC = \angle EHF$.

From the Inscribed Angle Theorem, $2 \angle BAC = \angle BGC$ and $2 \angle EDF = \angle EHF$.

Therefore $\angle BAC = \angle EHF$.

$\blacksquare$

Historical Note

This is Proposition 27 of Book III of Euclid's The Elements.

This theorem is the converse of Proposition 26: Equal Angles in Equal Circles.