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.

In the words of Euclid:

In equal circles equal angles stand on equal circumferences, whether at the center or at the circumferences.

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 $CG = FH$.

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 from Book $\text{III}$ Definition $11$: Similar Segments 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 proof is Proposition $26$ of Book $\text{III}$ of Euclid's The Elements.
It is the converse of Proposition $27$: Angles on Equal Arcs are Equal.