Angles on Equal Arcs are Equal

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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.


In the words of Euclid:

In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.

(The Elements: Book $\text{III}$: Proposition $27$)


Proof

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

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

Euclid-III-27.png

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 $2 \angle BAC = \angle EHF$.

$\blacksquare$


Historical Note

This theorem is Proposition $27$ of Book $\text{III}$ of Euclid's The Elements.
It is the converse of Proposition $26$: Equal Angles in Equal Circles.


Sources