Proof: By Euclid
(related to Proposition: 3.20: Inscribed Angle Theorem)
- For being joined, let $AE$ have been drawn through to $F$.
- Therefore, since $EA$ is equal to $EB$, angle $EAB$ (is) also equal to $EBA$ [Prop. 1.5].
- Thus, angle $EAB$ and $EBA$ is double (angle) $EAB$.
- And $BEF$ (is) equal to $EAB$ and $EBA$ [Prop. 1.32].
- Thus, $BEF$ is also double $EAB$.
- So, for the same (reasons), $FEC$ is also double $EAC$.
- Thus, the whole (angle) $BEC$ is double the whole (angle) $BAC$.
- So let another (straight line) have been inflected, and let there be another angle, $BDC$.
- And $DE$ being joined, let it have been produced to $G$.
- So, similarly, we can show that angle $GEC$ is double $EDC$, of which $GEB$ is double $EDB$.
- Thus, the remaining (angle) $BEC$ is double the (remaining angle) $BDC$.
- Thus, in a circle, the angle at the center is double that at the circumference, when [the angles] have the same circumference base.
- (Which is) the very thing it was required to show.
Thank you to the contributors under CC BY-SA 4.0!
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"