Equal Arcs of Circles Subtended by Equal Straight Lines

Theorem

In equal circles, equal arcs are subtended by equal straight lines.

Proof

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

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

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 arcs $BCG$ and $EHF$ are equal.

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

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 $EF$.

$\blacksquare$

Historical Note

This is Proposition 29 of Book III of Euclid's The Elements.

This theorem is the converse of Proposition 28: Straight Lines Cut Off Equal Arcs in Equal Circles.