Angles in Same Segment of Circle are Equal

Theorem

In the words of Euclid:

In a circle the angles in the same segment are equal to one another.

(The Elements: Book $\text{III}$: Proposition $21$)

Proof

Let $ABCD$ be a circle, and let $\angle BAD, \angle BED$ be angles in the same segment $BAED$.

Let $F$ be the center of $ABCD$, and join $BF$ and $FD$.

From the Inscribed Angle Theorem:

$\angle BFD = 2 \angle BAD$

$\angle BFD = 2 \angle BED$

So:

$\angle BAD = \angle BED$

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $21$ of Book $\text{III}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions
• 1991: David Wells: Curious and Interesting Geometry ... (previous) ... (next): angle in the same segment
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): circle $(1)$
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): circle $(1)$