# 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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