Angles in Circles have Same Ratio as Arcs
Theorem
In the words of Euclid:
- 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 $\text{VI}$: Proposition $33$)
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:
- $BC : EF = \angle BGC : \angle EHF = \angle BAC : \angle EDF$
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 Book $\text{V}$ Definition $5$: Equality of Ratios:
- $BC : EF = \angle BGL : \angle EHF$
But from the Inscribed Angle Theorem:
- $\angle BGC : \angle EHF = \angle BAC : \angle DEF$
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $33$ of Book $\text{VI}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions