Angles in Circles have Same Ratio as Arcs

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Theorem

As Euclid defined it:

In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences.

(The Elements: Book VI: Proposition $33$)

Let $ABC, DEF$ be equal circles.

Let $\angle BGC, \angle EHF$ be angles at their centres $G, H$.

Let $\angle BAC, \angle EDF$ be angles at their circumferences.

We need to show that $BC : EF = \angle BGC : \angle EHF = \angle BAC : \angle EDF$.

Let any number of consecutive arcs $CK, KL$ be made equal to the arc $BC$.

Also let any number of consecutive arcs $FM, MN$ be made equal to the arc $EF$.

Join $GK, GL, HM, HN$.

Then since $BC = CK = KL$ it follows from Angles on Equal Arcs are Equal that $\angle BGC = \angle CGK = \angle KGL$.

Therefore, whatever multiple the arc $BL$ is of $BC$, that multiple is also the angle $\angle BGL$ of the angle $\angle BGC$.

For the same reason, whatever multiple the arc $NE$ is of $EF$, that multiple is also the angle $\angle NHE$ of the angle $\angle EHF$.

$BL = EN \implies \angle BGL = \angle EHN$

$BL > EN \implies \angle BGL > \angle EHN$

$BL < EN \implies \angle BGL < \angle EHN$

Then from Book V Definition 5: Equality of Ratios $BC : EF = \angle BGL : \angle EHF$.

But $\angle BGC : \angle EHF = \angle BAC : \angle DEF$ from the Inscribed Angle Theorem.

Hence the result.

$\blacksquare$

This is Proposition 33 of Book VI of Euclid's The Elements.