circumferential angle is half the corresponding central angleCircumferentialAngleIsHalfCorrespondingCentralAngle2007-06-07 18:14:482007-12-14 20:57:39Theoremtheorems of Euclidcircumferential anglecentral angle% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thm}{Theorem}
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\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{ax}{Axiom}Consider a circle with center $O$ and two distinct points on the circle $A$ and $B$. If $C$ is a third point on the circle not equal to either $A$ or $B$, then the \emph{circumferential angle} at $C$ subtending the arc $AB$ is the angle $ACB$. Here, by arc $AB$, we mean the arc of the circle that does not contain the points $C$.
Similarly, the \emph{central angle} subtending arc $AB$ is the angle $AOB$. The central angle corresponds to the arc $AB$ measured on the same side of the circle as the angle itself. Note that if $AB$ is a diameter of the circle, then the central angle is $180^{\circ}$.
\begin{thm} \emph{[Euclid, Book III, Prop. 20]} In any circle, a circumferential angle is half the size of the central angle subtending the same arc.
\end{thm}
\begin{proof}
There are actually several distinct cases. Consider $\angle BAC$ in a circle with center $O$, and draw $AO, BO, CO$ as well as the chord containing both $A$ and $O$:
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%A B C triangle(4.2); draw(A,B,C)
%c=circle(A,B,C); draw(c)
%draw(A); draw("$A$",A,-180:)
%draw(B); draw("$B$",B,0:)
%draw(C); draw("$C$",C,90:)
%O=center(c); draw(O); draw("$O$",O,140:)
%F' F intersection(line(A,O),c)
%draw(segment(F',F))
%draw(F); draw("$F$",F,0:)
%a=segment(O,A)
%b=segment(O,B); draw(b)
%c=segment(O,C); draw(c)
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In this case, the center of the circle lies between the arms of the circumferential angle. Now, since $AO=OB$, $\triangle AOB$ is isosceles, and $\angle FOB$ is an exterior angle. Thus
\[\angle FOB=\angle OAB + \angle OBA = 2\angle OAB\]
Similarly, $\triangle AOC$ is isosceles, and
\[\angle FOC=\angle OAC + \angle OCA = 2\angle OAC\]
and it follows that
\[\angle BOC=\angle FOB + \angle FOC = 2\angle OAB + 2\angle OAC = 2\angle BAC\]
proving the result.
A second case is the case in which both arms of the angle lie to one side of the circle's center:
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%draw(B); draw("$B$",B,0:)
%draw(C); draw("$C$",C,90:)
%O=center(c); draw(O); draw("$O$",O,-45:)
%F' F intersection(line(A,O),c)
%draw(segment(F',F))
%draw(F); draw("$F$",F,0:)
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%b=segment(O,B); draw(b)
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\end{center}
The proof is similar to the previous case, except that the angle in question is the difference rather than the sum of two known angles. Here we see that both $\triangle AOB$ and $\triangle AOC$ are isosceles, so that again
\begin{align*}
\angle COF &= 2\angle OAC \\
\angle BOF &= 2\angle OAB
\end{align*}
Subtracting, we get
\[\angle COB = \angle COF - \angle BOF = 2\angle OAC -2\angle OAB = 2\angle BAC\]
as desired.
The final case is the case in which one arm of the angle goes through the center of the circle. This is a degenerate form of the first case, and the same proof follows through except that one of the angles is zero.
\end{proof}