Equal Arcs of Circles Subtended by Equal Straight Lines

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Theorem

In the words of Euclid:

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

(The Elements: Book $\text{III}$: Proposition $29$)


Proof

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

Let equal arcs $BGC$ 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 arcs $BGC$ 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 theorem is Proposition $29$ of Book $\text{III}$ of Euclid's The Elements.
It is the converse of Proposition $28$: Straight Lines Cut Off Equal Arcs in Equal Circles.


Sources