Straight Lines Cut Off Equal Arcs in Equal Circles

Theorem
In equal circles, equal straight lines cut off equal circumferences, 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 circumferences $ACB$ and $DFE$ as the greater, and $AGB, DHE$ as lesser.


 * Euclid-III-28.png

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 Equal Angles on Equal Circumferences the circumference $AGB$ equals the circumference $DHE$.

As the whole circles $ABC$ and $DEF$ are equal, the circumference $ACB$ which remains also equals circumference $DFE$.