Angles in Same Segment of Circle are Equal
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Theorem
In the words of Euclid:
(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