\section{Equal Angles in Equal Circles}
Tags: Circles

\begin{theorem}
In equal [[Definition:Circle|circles]], equal [[Definition:Angle|angles]] stand on equal [[Definition:Arc of Circle|arcs]], whether at the [[Definition:Center of Circle|center]] or at the [[Definition:Circumference of Circle|circumference]] of those circles.
{{:Euclid:Proposition/III/26}}
\end{theorem}

\begin{proof}
Let $ABC$ and $DEF$ be equal [[Definition:Circle|circles]].
Let $\angle BGC = \angle EHF$ and $\angle BAC = \angle EDF$.
:[[File:Euclid-III-26.png|500px]]
Let $BC$ and $EF$ be joined.
Since the [[Definition:Circle|circles]] $ABC$ and $DEF$ are equal, their [[Definition:Radius of Circle|radii]] are equal.
So $BG = EH$ and $CG = FH$.
We also have [[Definition:By Hypothesis|by hypothesis]] that $\angle BGC = \angle EHF$.
So from [[Triangle Side-Angle-Side Equality]] it follows that $BC = EF$.
Since $\angle BAC = \angle EDF$ we have from {{EuclidDefLink|III|11|Similar Segments}} that [[Definition:Segment of Circle|segment]] $BAC$ is [[Definition:Similar Segments|similar]] to [[Definition:Segment of Circle|segment]] $EDF$.
Moreover, these [[Definition:Segment of Circle|segments]] have equal [[Definition:Base of Segment|bases]].
So from [[Similar Segments on Equal Bases are Equal]], [[Definition:Segment of Circle|segment]] $BAC$ is equal to [[Definition:Segment of Circle|segment]] $EDF$. 
But as $ABC$ and $DEF$ are equal [[Definition:Circle|circles]], it follows that [[Definition:Arc of Circle|arc]] $BKC$ equals [[Definition:Arc of Circle|arc]] $ELF$.
{{qed}}
{{Euclid Note|26|III|{{EuclidNoteConverse|prop=27|title=Angles on Equal Arcs are Equal}}}}
\end{proof}


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