Angle at Centre is Twice Angle at Circumference

Statement

For points \(A,B,C\) on a circle and \(O\) the centre: the angle at the centre subtended by chord \(AB\) is twice the angle at the circumference on the same arc: \[ \angle AOB \;=\; 2\,\angle ACB. \] Here \(C\) is any point on the circumference on the arc of \(AB\) that does not contain the centre angle you are using (i.e. the corresponding arc). Any such \(C\) gives the same angle \( \angle ACB \).

Why it’s true (quick idea)

• The triangle \(AOB\) is isosceles (\(OA=OB\) are radii), so its base angles are equal.
• The angle at the circumference \( \angle ACB \) is an exterior angle to triangle \(AOB\) “seen” from the circle: it equals half the central angle standing on the same arc.
• Hence \( \angle AOB = 2\,\angle ACB \).

Spotting it in a diagram

Look for a chord \(AB\) and two types of angles built from it: (i) a central angle \( \angle AOB \) at the centre, and (ii) any angle at the circumference that subtends chord \(AB\) (for example \( \angle ACB \)). Those two are always in the ratio \(2:1\).

Special case: diameter

If \(AB\) is a diameter, then the central angle \( \angle AOB=180^\circ \). Therefore any angle subtending \(AB\) at the circumference is \[ \angle ACB = \tfrac{1}{2}\cdot 180^\circ = 90^\circ, \] so a triangle drawn on a diameter is a right triangle (Thales’ theorem).

Recipe

1. Identify the chord \(AB\) and the relevant central angle \( \angle AOB \) or the circumference angle \( \angle ACB \).
2. Use \( \angle AOB = 2\,\angle ACB \) to transfer the value (double or halve).
3. Finish with triangle angle sum, isosceles facts from radii, or cyclic quadrilateral facts as needed.

Common pairings

• Angles in the same segment are equal: all circumference angles subtending the same chord/arc are equal.
• Cyclic quadrilateral: opposite angles sum to \(180^\circ\).
• Radius facts: \(OA=OB=OC\) gives isosceles triangles and base-angle equalities.

Mini examples

1. If \( \angle AOB=120^\circ\), then \( \angle ACB=60^\circ\).
2. If \( \angle ACB=34^\circ\), then \( \angle AOB=68^\circ\).
3. If \(AB\) is a diameter, then any \( \angle ACB=90^\circ\).

Pitfalls

• Using the wrong arc: the circumference angle must subtend the same arc as the central angle you chose.
• Mixing with tangent results: the alternate segment theorem is a different equality (tangent–chord angle equals an angle in the opposite segment).
• Forgetting that all circumference angles on the same arc are equal—use this to move values around the circle quickly.

Summary

The angle at the centre is twice the angle at the circumference standing on the same arc. Use it to halve central angles, double circumference angles, and to get instant right angles when a diameter is involved.