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.∎