Circle angle properties

Draw any circle, mark a chord, and look at the angles that chord throws off to the centre and to the rest of the circle's edge — a beautiful rigidity shows up. However you slide a point along the remaining arc, the angle it sees the chord at never changes, and it is locked to exactly half of whatever angle the same chord makes at the centre. Push the chord all the way to a diameter and that half-angle collapses to a clean right angle every single time. These are the three circle angle-property theorems — central angle double the inscribed angle, equal angles in the same segment, and the angle in a semicircle — and together they are the workhorse tools of circle geometry: most "find the angle" problems in this chapter reduce, after a little angle-chasing, to spotting which of the three applies.

The core insight

The angle a chord subtends at the centre of a circle is always double the angle it subtends at any point on the remaining arc, and this single fact forces equal angles in the same segment and a right angle in a semicircle as special cases.

The core idea

1. Angle at the centre is double the angle at the circumference. Let arc $AB$ subtend $\angle AOB$ at the centre $O$ , and let $C$ be any point on the major arc (the remaining part of the circle not containing $A$ and $B$ 's minor arc). Then:

\angle AOB = 2\,\angle ACB

where: $O$ is the centre, $\angle AOB$ is the central angle standing on arc $AB$ , and $\angle ACB$ is the inscribed angle standing on the same arc $AB$ from any point $C$ on the remaining part of the circle. The word "same arc" matters: both angles must be looking at the identical arc $AB$ from opposite sides of it.

2. Angles in the same segment are equal. If $C$ and $D$ both lie on the same arc, on the same side of chord $AB$ , then the angles they subtend on $AB$ are equal:

\angle ACB = \angle ADB

This follows immediately from theorem 1: both $\angle ACB$ and $\angle ADB$ equal half of the same fixed central angle $\angle AOB$ , so they must equal each other. (This result is quoted "without proof" on the syllabus — you apply it directly.)

3. Angle in a semicircle is a right angle. If $AB$ is a diameter and $C$ is any point on the circle, then:

\angle ACB = 90^\circ

A diameter subtends a straight angle $180^\circ$ at the centre, so by theorem 1 any point on the circumference sees it at $\tfrac{180^\circ}{2} = 90^\circ$ . This is not a separate fact to memorise — it is theorem 1 with the central angle fixed at $180^\circ$ .

Figure

Circle with centre O, points A and B on the circle joined to O showing central angle AOB, and point C on the major arc joined to A and B showing inscribed angle ACB equal to half of AOB

Derivation

Derivation ACentral angle equals twice the inscribed angle

The goal is to prove, using nothing beyond the exterior-angle property of a triangle and the fact that two radii of the same circle are equal, that the angle a chord subtends at the centre is always exactly double the angle it subtends at any point on the remaining arc, so that this theorem can be relied on as proven rather than merely observed in diagrams.

Step 1. Draw circle with centre $O$ , chord $AB$ , and a point $C$ on the major arc. Join $C$ to $O$ and extend $CO$ to meet the circle again at point $D$ , so $CD$ passes through the centre. This construction splits both $\angle AOB$ (the angle we want) and $\angle ACB$ (the angle we compare it to) into two manageable pieces, $\angle AOD$ , $\angle BOD$ and $\angle ACD$ , $\angle BCD$ .

Step 2. Focus on triangle $OAC$ . Since $OA$ and $OC$ are both radii of the same circle, $OA = OC$ , so triangle $OAC$ is isosceles and $\angle OAC = \angle OCA$ . The angle $\angle AOD$ is the exterior angle of triangle $OAC$ at vertex $O$ (it lies along the straight line $CD$ ), so by the exterior-angle theorem $\angle AOD = \angle OAC + \angle OCA = 2\,\angle OCA = 2\,\angle ACD$ .

Step 3. Repeat the same argument for triangle $OBC$ . Here $OB = OC$ (radii again), so $\angle OBC = \angle OCB$ , and $\angle BOD$ is the exterior angle of triangle $OBC$ at $O$ , giving $\angle BOD = 2\,\angle OCB = 2\,\angle BCD$ .

Step 4. Add the two results. $\angle AOD + \angle BOD = 2\,\angle ACD + 2\,\angle BCD$ , which is $\angle AOB = 2(\angle ACD + \angle BCD) = 2\,\angle ACB$ . (If $O$ lies outside $\angle ACB$ instead of inside it, the same two steps give a subtraction, $\angle AOB = 2\,\angle BCD - 2\,\angle ACD$ , but the final relation $\angle AOB = 2\,\angle ACB$ still holds; the addition case above is the one you are expected to reproduce.) This proves the theorem for any position of $C$ on the major arc.

Worked example

$O$ is the centre of a circle, and $A$ , $B$ , $C$ are points on the circle such that $\angle AOB = 100^\circ$ , with $C$ on the major arc. Find $\angle ACB$ . Then, $PQ$ is a diameter of the same circle with $R$ on the circle such that $\angle RPQ = 35^\circ$ ; find $\angle PRQ$ and $\angle RQP$ .

Step 1. Identify the theorem for the first part. $\angle AOB$ is the central angle on arc $AB$ , and $\angle ACB$ is the inscribed angle on the same arc from $C$ on the remaining part of the circle, so theorem 1 applies directly.

Step 2. Apply $\angle AOB = 2\,\angle ACB$ : substituting $100^\circ = 2\,\angle ACB$ gives $\angle ACB = \dfrac{100^\circ}{2} = 50^\circ$ .

Step 3. For the second part, use the semicircle theorem. Since $PQ$ is a diameter, any point on the circle — including $R$ — sees it at a right angle, so $\angle PRQ = 90^\circ$ .

Step 4. Find $\angle RQP$ from the angle sum of triangle $PRQ$ : $\angle RQP = 180^\circ - \angle PRQ - \angle RPQ = 180^\circ - 90^\circ - 35^\circ = 55^\circ$ .

Final answer: $\angle ACB = 50^\circ$ ; $\angle PRQ = 90^\circ$ and $\angle RQP = 55^\circ$ .

Competitive stretch

If a second point $S$ also lies on the major arc of $PQ$ on the same side as $R$ , then by the same-segment theorem $\angle PSQ = \angle PRQ = 90^\circ$ automatically — no fresh calculation is needed, since $R$ and $S$ subtend the same chord $PQ$ from the same segment.

Common traps

Applying $\angle AOB = 2\,\angle ACB$ with $C$ on the wrong arc — the point must lie on the arc not containing the arc referred to by the central angle, i.e. the "remaining part of the circle."
Confusing "angles in the same segment" (equal) with angles in opposite segments, which are supplementary via the cyclic-quadrilateral property, not equal.
Treating the semicircle theorem as a separate fact to memorise, instead of recognising it as theorem 1 with the central angle fixed at $180^\circ$ .
Assuming any triangle drawn inside a circle has a right angle — this is only guaranteed when one side of the triangle is a diameter.

Ask the tutor: Want to practice spotting which of the three circle-angle theorems applies to a busy diagram — ask the AI tutor to walk through a few figures with you.

Why does C have to be on the major arc for angle AOB equals 2 times angle ACB to work?

The theorem compares the central angle on one arc with the inscribed angle from a point on the other arc. If C sat on the minor arc instead, it would be looking at the reflex angle at O, not the marked angle AOB, so the simple doubling relation would not hold in the form given. Always check which arc the central angle "belongs to" and pick your inscribed point from the remaining part of the circle.

Is the same-segment theorem something I need to prove in an exam?

No. Per the syllabus, "angles in the same segment are equal" is stated without proof — you are expected to know it and apply it to find angles, not derive it from scratch. The central-angle theorem is different: its proof is examinable, so make sure you can reproduce the exterior-angle argument shown above.

How is the semicircle theorem connected to the central angle theorem?

A diameter is just a chord that passes through the centre, so it subtends a straight angle of 180 degrees at the centre. Feeding that into angle AOB equals 2 times angle ACB gives angle ACB equals 90 degrees automatically. So you never need to memorise the semicircle result separately — it drops straight out of the central-angle theorem once you notice the central angle is a straight line.