E5.6 - Circle angle theorems (Extended)

E5.6 - Circle angle theorems (Extended)

The same two endpoints can define an acute angle on one side of a circle and an obtuse angle on the other. How do you choose the correct angle at the centre? E5.6 connects centre and circumference angles, angles in the same segment and opposite angles of cyclic quadrilaterals, always keeping track of the arc involved.

Centre and circumference

Choose two endpoints AA and BB on a circle. An angle ACBACB, with CC elsewhere on the circumference, looks across the arc from AA to BB that does not contain CC. That is the arc subtended by the angle.

Centre and circumference theorem

The angle at the centre is twice the angle at the circumference standing on the same arc. If α\alpha is the selected centre angle and β\beta is the circumference angle, then α=2β\alpha=2\beta. The centre angle may be reflex.

For example, if the minor arc ABAB has central angle 120∘120^\circ, a circumference vertex on the major arc gives angle 60∘60^\circ. Move the vertex onto the minor arc: the angle now stands on the major arc, whose centre angle is 360∘−120∘=240∘360^\circ-120^\circ=240^\circ. That circumference angle is 120∘120^\circ.

[DIAGRAM: asset_name: Selecting the same circle arc; asset_slug: centre_and_same_segment; caption: The selected arc determines whether to use the smaller or reflex centre angle; vertices in the same segment see the same arc.; description: Three exact panels use a chord with minor central angle 120 degrees. In the first, a vertex on the major arc gives a 60-degree circumference angle; in the second, a vertex on the minor arc gives a 120-degree circumference angle corresponding to the reflex 240-degree centre angle. The third places two vertices on the same side of the chord and marks their equal 60-degree angles. Highlighted arcs avoid the circumference vertices.]
The selected arc determines whether to use the smaller or reflex centre angle

The theorem comes from equal radii. Joining the centre to the circumference vertex splits the figure into isosceles triangles. Their equal base angles, the triangle sum and the full turn at the centre give the double-angle relationship. If the joining line falls outside the angle, subtract the two isosceles-triangle angle relationships instead. Tracking the selected arc keeps the same result in each position.

Worked example 1: halving a centre angle

Question. A circle has centre OO. Its smaller angle AOBAOB is 138∘138^\circ. Point CC lies on the major arc ABAB. Find ∠ACB\angle ACB, giving a reason.

Plan. Identify the arc not containing CC, then halve its centre angle.

selected arc=minor arc AB∠ACB=138∘÷2=69∘\begin{aligned} \text{selected arc}&=\text{minor arc }AB\\ \angle ACB&=138^\circ\div2=69^\circ \end{aligned}

Answer. 69∘69^\circ, because the centre angle is twice the circumference angle on the same arc.

Check. Doubling 69∘69^\circ returns 138∘138^\circ. Since the selected arc is smaller than a semicircle, the circumference angle is acute.

Now the circumference vertex lies on the shorter arc, so doubling gives a reflex centre angle first.

Worked example 2: selecting a reflex angle

Question. AA, BB and CC lie on a circle with centre OO. Point CC is on the minor arc ABAB, and ∠ACB=117∘\angle ACB=117^\circ. Find the smaller angle AOBAOB, giving reasons.

Plan. Double the circumference angle to find the reflex angle on its major arc, then subtract from a full turn.

reflex ∠AOB=2(117∘)=234∘centre angle on the same major arcsmaller ∠AOB=360∘−234∘=126∘angles around a point\begin{aligned} \text{reflex }\angle AOB&=2(117^\circ)=234^\circ &&\text{centre angle on the same major arc}\\ \text{smaller }\angle AOB&=360^\circ-234^\circ=126^\circ &&\text{angles around a point} \end{aligned}

Answer. The smaller angle AOBAOB is 126∘126^\circ.

Check. The two centre angles sum to 360∘360^\circ. The angle on the major arc must be reflex, agreeing with 234∘234^\circ, while the requested smaller angle is obtuse.

When the selected arc is a semicircle, the centre angle is 180∘180^\circ, giving 90∘90^\circ at the circumference. This includes the diameter theorem you already know.

Watch out

Do not automatically double an angle and report the result. First identify the two endpoints, then the arc, then whether the question asks for the smaller or reflex centre angle.

Use the position of the circumference vertex in the next calculation.

Same segment

The centre angle on an arc stays fixed. Any circumference angles standing on that same arc are both half the same centre angle, so they are equal.

Angles in the same segment

If CC and DD lie on the same side of chord ABAB, on the same circle and away from AA and BB, then ∠ACB=∠ADB\angle ACB=\angle ADB. These are angles in the same segment.

Check both endpoints of the angles' arms. The angles may look different because their vertices are far apart, but each must join to AA and BB. If the vertices are on opposite sides of the chord, this equality does not apply.

Worked example 3: transferring an angle

Question. A,B,C,DA,B,C,D lie on one circle. Points CC and DD are on the same side of chord ABAB. Given ∠ACB=46∘\angle ACB=46^\circ, find ∠ADB\angle ADB, giving a reason.

Plan. Compare the two endpoints and the positions of their vertices.

  1. Both angles have arms ending at AA and BB.
  2. Their vertices CC and DD are in the same segment.
  3. Therefore ∠ADB=∠ACB=46∘\angle ADB=\angle ACB=46^\circ.

Answer. 46∘46^\circ, because angles in the same segment are equal.

Check. Both angles correspond to the same 92∘92^\circ centre angle. They must therefore each be 46∘46^\circ.

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