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This page sets out the method for circle theorem proofs in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Circle theorem proofs topic page.
When you see 'work out angle … giving reasons' or 'prove that angle … is ', do this.
When you see 'prove that' with no numbers on the diagram, do this.
The angle at the centre and the angle at the edge must stand on the same arc before you double or halve.
The centre angle is twice the edge angle on that arc.
If you know the centre, the edge angle on the same arc is half.
Two radii make an isosceles triangle. The base angles match.
A proof answer names the fact you used, not only the size of the angle.
Find the isosceles angles first if you need them, then double or halve.
Let O be the centre, and let A, B and C lie on the circumference. You want angle AOC, the angle at the centre, and angle ABC, the angle at the circumference on the same arc AC. Draw the radius OB, and take the usual diagram in which OB falls inside angle ABC, so B sits on the major arc. Triangle AOB has OA = OB because both are radii, so it is isosceles and its base angles are equal. Call each of them x. Triangle COB has OC = OB, so it is isosceles as well. Call each base angle y. The circumference angle is the pair at B, so angle ABC = x + y.
Do not stop at the angles inside the two triangles. Those are angle AOB = 180^° - 2x and angle COB = 180^° - 2y, and neither of them is angle AOC. Extend BO past the centre to a point D, so B, O and D lie on a straight line. Angle AOD is the exterior angle of triangle AOB at O. An exterior angle equals the sum of the two opposite interior angles, which are both x, so angle AOD = 2x. In the same way angle COD = 2y. The ray OD lies inside angle AOC, so the centre angle is the sum of those exterior angles: angle AOC = 2x + 2y = 2(x + y) = 2angle ABC.
The letters stay general until you want a check. Suppose the two base angles on the circumference side are 28^° and 35^°. Then x = 28 and y = 35, the edge angle is 28 + 35 = 63^°, and the centre angle is 2x + 2y = 56 + 70 = 126^°. That is the same as 2(28 + 35) = 126. The doubling is the last sentence of the proof. The reasons in front of it are equal radii, isosceles base angles, and the exterior angle.
Method: 1. Draw radius OB and mark equal base angles x and y in the two isosceles triangles. 2. The circumference angle is x+y. 3. Extend BO through the centre and add the exterior angles, so the centre angle is 2(x+y).
Let AB be a diameter and let C be any other point on the circumference. The angle at the centre standing on arc AB is the angle on a straight line, because a diameter is a straight line through the centre. So that centre angle is 180^°. Angle ACB stands on the same arc. The theorem you have just proved says the circumference angle is half the centre angle, so angle ACB = 1/2 × 180^° = 90^°. This is why the angle in a semicircle is a right angle: it is the twice-angle result with the centre angle fixed at a half turn.
Once the right angle is known, the other two angles of triangle ABC come from the angle sum of a triangle. They add to 90^°. If angle CAB is 35^°, then angle ABC = 180^° - 90^° - 35^° = 55^°. The 90 must be named as the angle in a semicircle, or as half of 180, before you subtract. A triangle with a diameter has a right angle at the circumference, not at the centre.
Method: 1. The angle at the centre on a diameter is a straight angle, 180^°. 2. Halve it: the angle at the circumference is 1/2× 180^°=90^°. 3. The other two angles of the triangle then add to 90^°.
Let ABCD be a quadrilateral with all four vertices on one circle. Angle ABC stands on arc ADC. Angle ADC, the opposite angle, stands on the opposite arc ABC. From the centre theorem, an angle at the circumference is half the angle at the centre on the same arc, and that centre angle is the degree measure of the arc. So each of these angles is half of the arc it stands on. Arc ADC and arc ABC sit next to each other and together make the whole circle, which is 360^°. Therefore angle ABC + angle ADC = 1/2 × 360^° = 180^°. The same argument fits the other pair of opposite angles.
A numeric check uses the arcs rather than a fresh diagram. Suppose the opposite arcs are 160^° and 200^°. The inscribed angles are half of those arcs, so they are 80^° and 100^°. Their sum is 80 + 100 = 180. Half of 160 plus half of 200 is half of 360, which is why the sum is a straight angle even when the arcs are unequal. The reason to write is that the arcs make a full circle, not that "opposite angles look similar".
Method: 1. Each opposite angle is half the arc it stands on. 2. Those two arcs together make 360^°. 3. So the opposite angles add to 1/2× 360^°=180^°.
A marker can follow a proof when every line has one job. Start by naming equal lengths: OA, OB and OC are radii, so the two triangles are isosceles. Next name the equal base angles, and give them letters. Then say where the circumference angle sits, as x + y. Only after that should you extend the radius and quote the exterior-angle fact. The last line is the comparison angle AOC = 2(x + y). If you swap that order, the double appears before the reader knows what is being doubled.
Keep the semicircle and the cyclic quadrilateral as consequences, not as extra starting facts, when the question says "prove". For a semicircle, quote a straight angle at the centre and then halve it. For a cyclic quadrilateral, say which arc each opposite angle stands on, add the arcs to 360^°, and halve. Same-arc language has to be visible. An angle on a different arc is not half, and it is not double, of the angle you were given.
Method: 1. Name the equal radii first, then the base angles x and y. 2. Write the circumference angle as x+y before you double it. 3. For a cyclic quadrilateral, add the opposite arcs to 360^° and then halve.
Circle theorem questions come in three styles. The first says "work out the size of angle x" and then "give a reason for your answer". The second sets algebra in the angles, such as (3x + 4)^°, and asks you to find x. The third says "prove that" or "show that" and expects a chain of statements, each with a theorem as its reason. The diagram always says "not drawn accurately", so only use facts that are given or that a theorem allows.
A reason is a short, exact phrase. "Angle at the centre is twice the angle at the circumference" is a reason. "Because it is a circle" is not. Write each angle on the diagram as you find it, so the chain of working is visible. Marks are awarded for each correct angle with its reason, and a wrong early angle does not remove later marks if you follow it through correctly.
Method: 1. Mark every angle that is given, and any right angle from a tangent or a diameter. 2. Pick the next angle you can find and name the theorem that gives it. 3. Write the angle and the reason together, on one line. 4. Finish by answering the exact question, with units of degrees.
Tip: Write the reason as soon as you write the angle. Adding them later is how marks get lost.
Learn these exact phrases. The angle at the centre is twice the angle at the circumference. Angles in the same segment are equal. The angle in a semicircle is 90^°. Opposite angles of a cyclic quadrilateral add up to 180^°. A tangent is perpendicular to the radius at the point of contact. Tangents from an external point are equal in length. The angle between a tangent and a chord equals the angle in the alternate segment.
Two more help with chords. The perpendicular from the centre to a chord bisects the chord. Two radii of the same circle are equal, so a triangle with two radii as sides is isosceles and has equal base angles. These last two are the reasons that join a circle theorem to Pythagoras or to algebra. When you quote the reason, name the angles, for example "angle ACB is in a semicircle, so it is 90^°".
Method: 1. Decide which theorem matches the shape: centre, segment, semicircle, cyclic quadrilateral, tangent. 2. Quote the exact phrase for that theorem. 3. Name the angles you are using in the sentence. 4. Use an isosceles triangle with two radii if the centre is on the diagram.
Higher questions put expressions in the angles. If two opposite angles of a cyclic quadrilateral are (4x + 5)^° and (2x + 25)^°, they add up to 180^°, so 6x + 30 = 180 and x = 25. The angles are 105^° and 75^°. Always substitute back to find the angle the question asks for, because the question may want an angle and not x.
The same method works for the angle in a semicircle. If a triangle has a diameter as one side, the angle opposite is 90^°, and the other two angles add to 90^°. Set up the equation, simplify, solve, and then check that the three angles add to 180^° or the opposite pair to 180^°. State the theorem before the equation so the marker can see why it is true.
Method: 1. Write the theorem as an equation using the expressions in the diagram. 2. Collect like terms and solve for x. 3. Substitute back to find the required angle. 4. Check that the angles satisfy the theorem you used.
A proof starts by naming an unknown angle, such as "let angle CAB = x". Each following line gives a new angle in terms of x with a reason, and the last line states what was to be proved. Common starting points are isosceles triangles formed by two radii, a diameter that gives 90^°, or a tangent that gives 90^° with the radius. Join the points of contact to the centre if the diagram does not already show it.
Do not use the result you are trying to prove as one of your reasons. A proof that the opposite angles of a cyclic quadrilateral add to 180^° must use the centre-angle theorem, not the cyclic quadrilateral theorem itself. Finish with a closing sentence such as "so angle BCD = x, as required". When a question gives numbers as well, use them to check your algebra with a second method.
Method: 1. Let a convenient angle be x. 2. Find other angles in terms of x, giving a reason on every line. 3. Make sure no step uses the result being proved. 4. End with the statement that was to be proved.
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