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.^{1}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes