Straight Lines Cut Off Equal Arcs in Equal Circles
Theorem
In the words of Euclid:
- In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less.
(The Elements: Book $\text{III}$: Proposition $28$)
Proof
Let $ABC$ and $DEF$ be equal circles.
Let $AB, DE$ be equal straight lines cutting off arcs $ACB$ and $DFE$ as the greater, and $AGB, DHE$ as lesser.
Let $K$ and $L$ be the centers of the circles $ABC$ and $DEF$ respectively.
Let $AK, KB, DL, LE$ be joined.
Since the circles are equal, so are their radii.
So $AK = DL, KB = LE$.
As $AB = DE$ by hypothesis, from Triangle Side-Side-Side Equality it follows that $\angle AKB = \angle DLE$.
But from Angles on Equal Arcs are Equal the arc $AGB$ equals the arc $DHE$.
As the whole circles $ABC$ and $DEF$ are equal, the arc $ACB$ which remains also equals arc $DFE$.
$\blacksquare$
Historical Note
This proof is Proposition $28$ of Book $\text{III}$ of Euclid's The Elements.
It is the converse of Proposition $29$: Equal Arcs of Circles Subtended by Equal Straight Lines.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions