Equal Angles in Equal Circles

Theorem

In equal circles, equal angles stand on equal arcs, whether at the center or at the circumference of those circles.

Proof

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

Let $\angle BGC = \angle EHF$ and $\angle BAC = \angle EDF$.

Let $BC$ and $EF$ be joined.

Since the circles $ABC$ and $DEF$ are equal, their radii are equal.

So $BG = EH$ and $DC = HF$.

We also have by hypothesis that $\angle BGC = \angle EHF$.

So from Triangle Side-Angle-Side Equality it follows that $BC = EF$.

Since $\angle BAC = \angle EDF$ we have by definition that segment $BAC$ is similar to segment $EDF$.

Moreover, these segments have equal bases.

So from Similar Segments on Equal Bases are Equal, segment $BAC$ is equal to segment $EDF$.

But as $ABC$ and $DEF$ are equal circles, it follows that arc $BKC$ equals arc $ELF$.

$\blacksquare$

Historical Note

This is Proposition 26 of Book III of Euclid's The Elements.

This theorem is the converse of Proposition 27: Angles on Equal Arcs are Equal.