Angles on Equal Arcs are Equal

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 $EF$.

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 proof 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

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions