Straight Lines Cut Off Equal Arcs in Equal Circles

Theorem

In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater and the less equal to the less.

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 is Proposition 28 of Book III of Euclid's The Elements.

This theorem is the converse of Proposition 29: Equal Arcs of Circles Subtended by Equal Straight Lines.