Angles in Circles have Same Ratio as Arcs

Proof
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:
 * $BC : EF = \angle BGC : \angle EHF = \angle BAC : \angle EDF$


 * Euclid-VI-33.png

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 $\angle EHF$.

Then from Angles on Equal Arcs are Equal:
 * $BL = EN \implies \angle BGL = \angle EHN$
 * $BL > EN \implies \angle BGL > \angle EHN$
 * $BL < EN \implies \angle BGL < \angle EHN$

Then from :
 * $BC : EF = \angle BGL : \angle EHF$

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

Hence the result.