circumferential angle is half the corresponding central angle

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 $\mathrm{\angle }BAC$ in a circle with center $O$, and draw $AO,BO,CO$ as well as the chord containing both $A$ and $O$:

In this case, the center of the circle lies between the arms of the circumferential angle. Now, since $AO=OB$, $\mathrm{△}AOB$ is isosceles, and $\mathrm{\angle }FOB$ is an exterior angle. Thus

$$\mathrm{\angle }FOB=\mathrm{\angle }OAB+\mathrm{\angle }OBA=2\mathrm{\angle }OAB$$

Similarly, $\mathrm{△}AOC$ is isosceles, and

$$\mathrm{\angle }FOC=\mathrm{\angle }OAC+\mathrm{\angle }OCA=2\mathrm{\angle }OAC$$

and it follows that

$$\mathrm{\angle }BOC=\mathrm{\angle }FOB+\mathrm{\angle }FOC=2\mathrm{\angle }OAB+2\mathrm{\angle }OAC=2\mathrm{\angle }BAC$$

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:

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 $\mathrm{△}AOB$ and $\mathrm{△}AOC$ are isosceles, so that again

$\mathrm{\angle }COF$ $=2\mathrm{\angle }OAC$

$\mathrm{\angle }BOF$ $=2\mathrm{\angle }OAB$

Subtracting, we get

$$\mathrm{\angle }COB=\mathrm{\angle }COF-\mathrm{\angle }BOF=2\mathrm{\angle }OAC-2\mathrm{\angle }OAB=2\mathrm{\angle }BAC$$

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. ∎

Title	circumferential angle is half the corresponding central angle
Canonical name	CircumferentialAngleIsHalfTheCorrespondingCentralAngle
Date of creation	2013-03-22 17:13:28
Last modified on	2013-03-22 17:13:28
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	14
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 51M04
Related topic	AngleOfViewOfALineSegment
Related topic	RiemannSphere
Defines	circumferential angle
Defines	central angle