# 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