325. The Centre's Double · angle at centre is twice angle at circumference

Try to break it

Drag C around the major arc — the central angle ∠AOB always reads exactly 2 × ∠ACB. Drag C onto the minor arc and the relationship still holds, but the relevant central angle is the reflex one (∠AOB + reflex = 360°). The "angle at the centre is twice the angle at the circumference" never breaks; just watch which arc you're on.

How you build it

Build a chord with its central and inscribed angles, and see ∠AOB = 2·∠ACB.

• Point tool: mark O near the middle — this is the centre of the circle.
• Circle tool: click O as the centre, then click outward to set the radius and draw the circle.
• Point tool: mark A on the circle.
• Point tool: mark B elsewhere on the circle — AB is the chord (the arc it cuts off carries both angles).
• Point tool: mark C on the circle on the major (longer) arc AB, on the far side of the chord from O.
• Segment tool: join A to B — the chord that both angles stand on.
• Segment tool: join O to A.
• Segment tool: join O to B. Now ∠AOB is the angle the arc makes at the centre.
• Segment tool: join C to A.
• Segment tool: join C to B. Now ∠ACB is the angle at the circumference. On the same arc AB, the central angle is always twice it: ∠AOB = 2·∠ACB.

The proof, step by step

Prove that the angle at the centre is twice the angle at the circumference on the same arc.

• Join OC and extend it to D. In ∆OAC, OA = OC (radii) ⇒ ∠OAC = ∠OCA. Exterior ∠AOD = ∠OAC + ∠OCA = 2∠OCA.
• In ∆OBC, OB = OC (radii) ⇒ ∠OBC = ∠OCB. Exterior ∠BOD = ∠OBC + ∠OCB = 2∠OCB.
• ∠AOB = ∠AOD + ∠BOD = 2∠OCA + 2∠OCB = 2(∠OCA + ∠OCB) = 2∠ACB.

Worked example

In a circle with centre O, points A, B, and C lie on the circumference. If ∠ACB = 35°, what is ∠AOB?

By the Angle at Centre theorem, ∠AOB = 2∠ACB = 2 × 35° = 70°.

• 17.5°
• 35°
• 70° — correct
• 140°