Equal Arcs of Circles Subtended by Equal Straight Lines

Theorem
In equal circles, equal circumferences are subtended by equal straight lines.

Proof
Let $ABC$ and $DEF$ be equal circles.

Let equal circumferences $BCG$ and $EHF$ be cut off by the straight lines $BC$ and $EF$.


 * Euclid-III-29.png

Let $K$ and $L$ be the centers of the circles $ABC$ and $DEF$ respectively.

Let $BK, KC, EL, LF$ be joined.

We have that the circumferences $BCG$ and $EHF$ are equal.

So from Equal Angles on Equal Circumferences $\angle BKC = \angle ELF$.

Since circles $ABC$ and $DEF$ are equal, so are their radii.

So $BK = EL$ and $KC = LF$, and they contain equal angles.

So from Triangle Side-Angle-Side Equality base $BC$ is equal to base $EF$.