Equal Angles in Equal Circles

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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.

(The Elements: Book $\text{III}$: Proposition $26$)


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 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 theorem 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.


Sources