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 $\angle BAC = \angle EDF$.


 * Euclid-III-26.png

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 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 $ELF$.