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[parent] circumferential angle is half the corresponding central angle (Theorem)

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 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 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 $180^{\circ}$ .

Theorem 1   [Euclid, Book III, Prop. 20] In any circle, a circumferential angle is half the size of the central angle subtending the same arc.
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 $O$ :

\begin{pspicture*}(-1.0000,-2.0000)(5.0000,3.2000) \rput(-1.1,-2){.} \rput(5,4){... ...36,0.3873)(3.1164,0.0949) \psline(1.6920,1.4906)(1.9830,1.5637) \end{pspicture*}

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

$\displaystyle \angle FOB=\angle OAB + \angle OBA = 2\angle OA$
Similarly, $ \triangle AOC$ is isosceles, and

$\displaystyle \angle FOC=\angle OAC + \angle OCA = 2\angle OA$
and it follows that

$\displaystyle \angle BOC=\angle FOB + \angle FOC = 2\angle OAB + 2\angle OAC = 2\angle BA$
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:


\begin{pspicture*}(-1.0000,-3.0000)(5.0000,2.0000) \rput(-1.1,-3){.} \rput(5,2.1... ...,-0.2412)(3.2013,-0.5231) \psline(2.1463,0.3621)(2.4418,0.3100) \end{pspicture*}
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
$\displaystyle \angle COF$ $\displaystyle = 2\angle OAC$    
$\displaystyle \angle BOF$ $\displaystyle = 2\angle OAB$    

Subtracting, we get$$\angle COB = \angle COF - \angle BOF = 2\angle OAC -2\angle OAB = 2\angle BA$$ 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. $ \qedsymbol$




"circumferential angle is half the corresponding central angle" is owned by rm50.
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See Also: angle of view of a line segment, Riemann sphere

Also defines:  circumferential angle, central angle

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Cross-references: sum, difference, similar, proof, exterior angle, isosceles, chord, size, diameter, side, contain, mean, angle, arc, points, center, circle
There are 14 references to this entry.

This is version 11 of circumferential angle is half the corresponding central angle, born on 2007-06-07, modified 2007-12-14.
Object id is 9551, canonical name is CircumferentialAngleIsHalfCorrespondingCentralAngle.
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Classification:
AMS MSC51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)

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note on circumferencial/central angle by perucho on 2007-12-15 06:50:23
Dear Roger,
In your interesting entry one also may obtain intrinsecally the subject's relation without reference to a fourth point F. The three internal radii bisect angles A,B,C according the equation
(\alpha+\gamma)+(\alpha+\beta)+(\beta+\gamma)=\pi,
so any out of the three central angles may be plugged from this. For example,
\pi-2\beta=2(\alpha+\gamma).
I would want to do such observation for if you want to add it.
Greetings,
Pedro
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Display issue by rm50 on 2007-06-07 18:17:53
Perhaps someone can tell me if this entry displays properly for you. For me, it has no diagrams and has very odd text displayed instead in HTML mode. In page image mode, it looks OK, except that the last several lines of the entry have disappeared. Thanks.
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