circumferential angle is half the corresponding central angle (original) (raw)
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∘.
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 ∠BAC in a circle with center O, and draw AO,BO,CO as well as the chord containing both A and O:
..ABCOF
In this case, the center of the circle lies between the arms of the circumferential angle. Now, since AO=OB, △AOB is isosceles, and ∠FOB is an exterior angle. Thus
| ∠FOB=∠OAB+∠OBA=2∠OAB |
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Similarly, △AOC is isosceles, and
| ∠FOC=∠OAC+∠OCA=2∠OAC |
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and it follows that
| ∠BOC=∠FOB+∠FOC=2∠OAB+2∠OAC=2∠BAC |
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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:
..ABCOF
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 △AOB and △AOC are isosceles, so that again
| ∠COF | =2∠OAC |
|---|---|
| ∠BOF | =2∠OAB |
Subtracting, we get
| ∠COB=∠COF-∠BOF=2∠OAC-2∠OAB=2∠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. ∎