# 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 $AO=OB$, $\mathrm{\u25b3}AOB$ 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, $\mathrm{\u25b3}AOC$ 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 $\mathrm{\u25b3}AOB$ and $\mathrm{\u25b3}AOC$ 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 |