GCSE Maths: Problems, algebra and proof (Higher)
Part of the GCSE Maths circle theorems chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Problems, algebra and proof
Exam questions rarely use one theorem on its own. Most ask you to chain two or three facts, add algebra, or prove a result. This section gives you a method for them.
A method for circle theorem problems
- 1. Mark the diagram. Write every angle you are told on it. Mark equal radii with ticks and mark the centre.
- 2. Look for the clues. Centre or radii (isosceles triangle, centre theorem). Diameter (90°). Tangent (90° with the radius, equal tangents, alternate segment). Four points on a circle (cyclic quadrilateral, same segment).
- 3. Find one new angle at a time. Write it with its reason in the standard wording, then use it for the next step.
- 4. If there is algebra, write the angle fact as an equation (equal angles: expression = expression; adding to 180°: sum = 180), solve, then substitute back.
- 5. Check. All angles in a triangle sum to 180°; try a second route if you can.
Show that and prove. In a “show that” or “prove” question the answer is given, so every step and its reason must be written down. A good layout is a list of statements, each with a reason: “angle OAB = 90° − x (base angle of an isosceles triangle …)”. Use letters instead of numbers so the proof is true for every case. End with a clear statement of what you have shown. Marks go for the method, the reasons and the conclusion. Do not use the result you are trying to prove.
Exam tip
Always write the reason next to the angle, in the standard words from Section 26.1. A correct angle with no reason scores fewer marks than a correct angle with a reason.Watch out
Do not use “it looks like” or measure the diagram. Diagrams in the exam are not accurate. Only use what you have been told or what you can prove from the theorems.- Work out angle OAB
- Work out angle AOB
Part (a)
Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.
- Find the right angle. Radius OA meets the tangent TU at the point of contact A.Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle OAU is made of angle OAB and angle BAU.Angle OAB = 90° − 38° = 52°
Check. 52° + 38° = 90°. ✓
Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.
Part (b)
Method. Triangle OAB has two equal sides (both radii), so it is isosceles.
- Write the base angles. OA and OB are radii, so the base angles are equal.Angle OBA = angle OAB = 52° (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the two base angles from 180°.Angle AOB = 180° − 52° − 52° = 76° (the angles in a triangle add up to 180°)
Check. Alternate segment theorem: the angle in the alternate segment is 38°, and the angle at the centre is 2 × 38° = 76°. ✓
Watch out. The apex angle of the isosceles triangle is at the centre O.
- Work out angle CAB.
- Work out angle ACB.
- Work out angle ABC.
Part (a)
Method. A tangent is perpendicular to the radius at the point of contact. AB is a diameter, so it contains the radius OA.
- Find angle UAB. The diameter AB is along the radius OA, and the tangent TU touches the circle at A.Angle UAB = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle UAB is made of angle UAC and angle CAB side by side.Angle CAB = 90° − 34° = 56°
Check. 34° + 56° = 90°. ✓
Watch out. The right angle is between the tangent and the diameter, so it contains both angles.
Part (b)
Method. AB is a diameter, so angle ACB is an angle in a semicircle.
- Spot the diameter. The two arms CA and CB go to the two ends of the diameter AB.Angle ACB = 90° (angle in a semicircle is 90°)
- Check the triangle. Triangle ABC now has a right angle at C.angle ACB = 90°
Check. Angle CAB + angle ACB = 56° + 90° = 146°, which is less than 180°, so there is room for angle ABC. ✓
Watch out. The right angle is at C, on the circle.
Part (c)
Method. Use the angle sum in triangle ABC, then check with the alternate segment theorem.
- Use the angles we know. Angle CAB is from part (a) and angle ACB is from part (b).Angle ABC = 180° − 56° − 90° = 34° (the angles in a triangle add up to 180°)
- Check with the alternate segment theorem. Angle UAC is between the tangent and the chord AC. The angle in the alternate segment on the other side of AC is angle ABC.Angle ABC = angle UAC = 34° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Both methods give 34°. ✓
Watch out. When two different routes give the same answer, you can be confident it is right.
- Work out angle ACB.
- Work out angle BAD.
- Work out angle BCD.
Part (a)
Method. Use the alternate segment theorem with the tangent PQ and the chord AB.
- Find the angle between the tangent and the chord. Angle QAB is between the tangent AQ and the chord AB.angle QAB = 50°
- Find the alternate segment. The alternate segment is on the other side of the chord AB, where C and D are. So the angle at C matches.Angle ACB = angle QAB = 50° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle ADB is also 50° (same segment), so both corners agree. ✓
Watch out. The alternate segment is the one on the opposite side of the chord from the tangent angle.
Part (b)
Method. P, A and Q are on a straight line (the tangent), so the three angles at A add up to 180°.
- Write the three angles at A. They are angle PAD, angle DAB and angle BAQ on the line PQ.45° + angle BAD + 50° = 180°
- Subtract. Angle BAD = 180° − 45° − 50° = 85° (angles on a straight line add up to 180°)
Check. 45 + 85 + 50 = 180. ✓
Watch out. The angle between the two chords is the middle angle on the straight line.
Part (c)
Method. ABCD is a cyclic quadrilateral and A is opposite C.
- Use the answer to part (b). Angle BAD is the interior angle at A.Angle BAD = 85°
- Use the cyclic quadrilateral rule. Opposite angles of a cyclic quadrilateral add up to 180°.Angle BCD = 180° − 85° = 95° (opposite angles of a cyclic quadrilateral add up to 180°)
Check. 85° + 95° = 180°. ✓
Watch out. Angle BCD is opposite angle BAD, not next to it.
- Work out angle BOC.
- Work out angle OBC.
- Work out angle ABC.
- Work out angle ABO.
Part (a)
Method. BAC is at the circumference and BOC is at the centre. Both stand on the arc BC.
- Name the arc. Both angles are made from the points B and C.arc BC
- Double it. The angle at the centre is twice the angle at the circumference.Angle BOC = 2 × 40° = 80° (the angle at the centre is twice the angle at the circumference)
Check. Half of 80° is 40°. ✓
Watch out. Use the circumference angle you are given and double it.
Part (b)
Method. OB and OC are radii, so triangle OBC is isosceles.
- Spot the isosceles triangle. Radii are equal, so the base angles are equal.angle OBC = angle OCB (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the angle at the centre from 180° and halve.Angle OBC = (180° − 80°) ÷ 2 = 50° (the angles in a triangle add up to 180°)
Check. 50° + 50° + 80° = 180°. ✓
Watch out. The apex of this triangle is at the centre O.
Part (c)
Method. Triangle ABC is isosceles because AB = AC.
- Write the base angles. The equal sides are AB and AC, so the base angles at B and C are equal.angle ABC = angle ACB (the base angles of an isosceles triangle are equal)
- Use the angle sum. The angle at A is 40°.Angle ABC = (180° − 40°) ÷ 2 = 70° (the angles in a triangle add up to 180°)
Check. 70° + 70° + 40° = 180°. ✓
Watch out. Here the apex is at A, on the circle, not at the centre.
Part (d)
Method. Angle ABC is made of angle ABO and angle OBC side by side.
- Write the sum. The radius BO lies inside angle ABC.angle ABO + angle OBC = angle ABC
- Subtract. Use angle ABC from part (c) and angle OBC from part (b).Angle ABO = 70° − 50° = 20°
Check. Triangle OAB is isosceles with OA = OB, so angle OAB = angle OBA = 20°, and AO bisects angle BAC = 40°. ✓
Watch out. Subtract the smaller part from the whole angle.
Method. Use the right angles at the points of contact in the quadrilateral OATB, then the congruent triangles.
- Find the angles at A and B. A tangent is perpendicular to the radius at the point of contact.angle OAT = angle OBT = 90° (the angle between a tangent and a radius is 90°)
- Use the angle sum of the quadrilateral. The angles in a quadrilateral add up to 360°.angle AOB + angle ATB + 90° + 90° = 360°
- Rearrange. Subtract 180° from both sides.angle AOB + angle ATB = 180°, so angle ATB = 180° − angle AOB. ✓
- Now show OT bisects angle ATB. Triangles OAT and OBT have a right angle, a common hypotenuse OT and OA = OB (radii).They are congruent (RHS).
- Match the angles at T. Congruent triangles have equal matching angles.angle ATO = angle BTO, so OT bisects angle ATB. ✓
Check. Test: if angle ATB = 64°, then angle AOB = 116° and each half at T is 32°. 64 + 116 = 180. ✓
Watch out. Use the standard reason after each line, for example “angles in a quadrilateral add up to 360°”.
Exercise 26J Multi-step circle theorem problems
Non-calculatorMethod. A tangent is perpendicular to the radius at the point of contact. AB is a diameter, so it contains the radius OA.
- Find angle UAB. The diameter AB is along the radius OA, and the tangent TU touches the circle at A.Angle UAB = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle UAB is made of angle UAC and angle CAB side by side.Angle CAB = 90° − 28° = 62°
Check. 28° + 62° = 90°. ✓
Watch out. The right angle is between the tangent and the diameter, so it contains both angles.
Method. AB is a diameter, so angle ACB is an angle in a semicircle.
- Spot the diameter. The two arms CA and CB go to the two ends of the diameter AB.Angle ACB = 90° (angle in a semicircle is 90°)
- Check the triangle. Triangle ABC now has a right angle at C.angle ACB = 90°
Check. Angle CAB + angle ACB = 62° + 90° = 152°, which is less than 180°, so there is room for angle ABC. ✓
Watch out. The right angle is at C, on the circle.
Method. Use the angle sum in triangle ABC, then check with the alternate segment theorem.
- Use the angles we know. Angle CAB is from part (a) and angle ACB is from part (b).Angle ABC = 180° − 62° − 90° = 28° (the angles in a triangle add up to 180°)
- Check with the alternate segment theorem. Angle UAC is between the tangent and the chord AC. The angle in the alternate segment on the other side of AC is angle ABC.Angle ABC = angle UAC = 28° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Both methods give 28°. ✓
Watch out. When two different routes give the same answer, you can be confident it is right.
Method. A tangent is perpendicular to the radius at the point of contact. AB is a diameter, so it contains the radius OA.
- Find angle UAB. The diameter AB is along the radius OA, and the tangent TU touches the circle at A.Angle UAB = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle UAB is made of angle UAC and angle CAB side by side.Angle CAB = 90° − 50° = 40°
Check. 50° + 40° = 90°. ✓
Watch out. The right angle is between the tangent and the diameter, so it contains both angles.
Method. AB is a diameter, so angle ACB is an angle in a semicircle.
- Spot the diameter. The two arms CA and CB go to the two ends of the diameter AB.Angle ACB = 90° (angle in a semicircle is 90°)
- Check the triangle. Triangle ABC now has a right angle at C.angle ACB = 90°
Check. Angle CAB + angle ACB = 40° + 90° = 130°, which is less than 180°, so there is room for angle ABC. ✓
Watch out. The right angle is at C, on the circle.
Method. Use the angle sum in triangle ABC, then check with the alternate segment theorem.
- Use the angles we know. Angle CAB is from part (a) and angle ACB is from part (b).Angle ABC = 180° − 40° − 90° = 50° (the angles in a triangle add up to 180°)
- Check with the alternate segment theorem. Angle UAC is between the tangent and the chord AC. The angle in the alternate segment on the other side of AC is angle ABC.Angle ABC = angle UAC = 50° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Both methods give 50°. ✓
Watch out. When two different routes give the same answer, you can be confident it is right.
Method. Use the alternate segment theorem with the tangent PQ and the chord AB.
- Find the angle between the tangent and the chord. Angle QAB is between the tangent AQ and the chord AB.angle QAB = 45°
- Find the alternate segment. The alternate segment is on the other side of the chord AB, where C and D are. So the angle at C matches.Angle ACB = angle QAB = 45° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle ADB is also 45° (same segment), so both corners agree. ✓
Watch out. The alternate segment is the one on the opposite side of the chord from the tangent angle.
Method. P, A and Q are on a straight line (the tangent), so the three angles at A add up to 180°.
- Write the three angles at A. They are angle PAD, angle DAB and angle BAQ on the line PQ.45° + angle BAD + 45° = 180°
- Subtract. Angle BAD = 180° − 45° − 45° = 90° (angles on a straight line add up to 180°)
Check. 45 + 90 + 45 = 180. ✓
Watch out. The angle between the two chords is the middle angle on the straight line.
Method. ABCD is a cyclic quadrilateral and A is opposite C.
- Use the answer to part (b). Angle BAD is the interior angle at A.Angle BAD = 90°
- Use the cyclic quadrilateral rule. Opposite angles of a cyclic quadrilateral add up to 180°.Angle BCD = 180° − 90° = 90° (opposite angles of a cyclic quadrilateral add up to 180°)
Check. 90° + 90° = 180°. ✓
Watch out. Angle BCD is opposite angle BAD, not next to it.
Method. Use the alternate segment theorem with the tangent PQ and the chord AB.
- Find the angle between the tangent and the chord. Angle QAB is between the tangent AQ and the chord AB.angle QAB = 40°
- Find the alternate segment. The alternate segment is on the other side of the chord AB, where C and D are. So the angle at C matches.Angle ACB = angle QAB = 40° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle ADB is also 40° (same segment), so both corners agree. ✓
Watch out. The alternate segment is the one on the opposite side of the chord from the tangent angle.
Method. P, A and Q are on a straight line (the tangent), so the three angles at A add up to 180°.
- Write the three angles at A. They are angle PAD, angle DAB and angle BAQ on the line PQ.35° + angle BAD + 40° = 180°
- Subtract. Angle BAD = 180° − 35° − 40° = 105° (angles on a straight line add up to 180°)
Check. 35 + 105 + 40 = 180. ✓
Watch out. The angle between the two chords is the middle angle on the straight line.
Method. ABCD is a cyclic quadrilateral and A is opposite C.
- Use the answer to part (b). Angle BAD is the interior angle at A.Angle BAD = 105°
- Use the cyclic quadrilateral rule. Opposite angles of a cyclic quadrilateral add up to 180°.Angle BCD = 180° − 105° = 75° (opposite angles of a cyclic quadrilateral add up to 180°)
Check. 105° + 75° = 180°. ✓
Watch out. Angle BCD is opposite angle BAD, not next to it.
Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.
- Find the right angle. Radius OA meets the tangent TU at the point of contact A.Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle OAU is made of angle OAB and angle BAU.Angle OAB = 90° − 24° = 66°
Check. 66° + 24° = 90°. ✓
Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.
Method. Triangle OAB has two equal sides (both radii), so it is isosceles.
- Write the base angles. OA and OB are radii, so the base angles are equal.Angle OBA = angle OAB = 66° (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the two base angles from 180°.Angle AOB = 180° − 66° − 66° = 48° (the angles in a triangle add up to 180°)
Check. Alternate segment theorem: the angle in the alternate segment is 24°, and the angle at the centre is 2 × 24° = 48°. ✓
Watch out. The apex angle of the isosceles triangle is at the centre O.
Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.
- Find the right angle. Radius OA meets the tangent TU at the point of contact A.Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle OAU is made of angle OAB and angle BAU.Angle OAB = 90° − 52° = 38°
Check. 38° + 52° = 90°. ✓
Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.
Method. Triangle OAB has two equal sides (both radii), so it is isosceles.
- Write the base angles. OA and OB are radii, so the base angles are equal.Angle OBA = angle OAB = 38° (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the two base angles from 180°.Angle AOB = 180° − 38° − 38° = 104° (the angles in a triangle add up to 180°)
Check. Alternate segment theorem: the angle in the alternate segment is 52°, and the angle at the centre is 2 × 52° = 104°. ✓
Watch out. The apex angle of the isosceles triangle is at the centre O.
Method. AC is a diameter, so angle ABC is an angle in a semicircle.
- Spot the diameter. Angle ABC is made at B on the circle by lines to the two ends of the diameter AC.AC is a diameter and B is on the circle
- Use the theorem. An angle made at the circle by the two ends of a diameter is a right angle.Angle ABC = 90° (angle in a semicircle is 90°)
Check. The triangle ABC then has angles 28°, 90° and 62°, which add to 180°. ✓
Watch out. The right angle is at B because B is the point on the circle.
Method. Use the angle sum of triangle ABC.
- Use the angles we know. Angle BAC is given and angle ABC is 90° from part (a).Angle BAC = 28° and angle ABC = 90°
- Subtract from 180°. The three angles of a triangle add up to 180°.Angle ACB = 180° − 90° − 28° = 62° (the angles in a triangle add up to 180°)
Check. 28° + 90° + 62° = 180°. ✓
Watch out. Subtract both known angles, not just one.
Method. Angles in the same segment are equal. Find two angles made from the same chord.
- Find the chord. Angle ADB and angle ACB are both made from the chord AB. D and C are on the same side of AB.Both angles stand on chord AB.
- Write the equal angle. Angles in the same segment are equal.Angle ADB = angle ACB = 62° (angles in the same segment are equal)
Check. Angle ADB + angle BAC + ... is not needed; instead angle at the centre AOB = 2 × 62° = 124°, and arc AB matches 2 × 62°. ✓
Watch out. "Same segment" means the same chord and the same side of it.
Method. Angle BCD is made of two angles side by side at C.
- Split the angle. The diagonal CA splits angle BCD into angle BCA and angle ACD.Angle BCD = angle BCA + angle ACD
- Add. We know both parts.Angle BCD = 62° + 35° = 97°
Check. Opposite angles BAD and BCD should add to 180°; see part (e). ✓
Watch out. Add the parts that are side by side. Do not subtract.
Method. ABCD is a cyclic quadrilateral, so opposite angles add up to 180°.
- Spot the opposite corners. A and C are opposite corners, so angle BAD is opposite angle BCD.Angle BCD = 97° (part d)
- Subtract from 180°. Opposite angles of a cyclic quadrilateral add up to 180°.Angle BAD = 180° − 97° = 83° (opposite angles of a cyclic quadrilateral add up to 180°)
Check. Angle BAD = angle BAC + angle CAD = 28° + 55° = 83° (triangle ACD has a right angle at D by the semicircle theorem). ✓
Watch out. Check that the two opposite angles really are opposite.
Method. AC is a diameter, so angle ABC is an angle in a semicircle.
- Spot the diameter. Angle ABC is made at B on the circle by lines to the two ends of the diameter AC.AC is a diameter and B is on the circle
- Use the theorem. An angle made at the circle by the two ends of a diameter is a right angle.Angle ABC = 90° (angle in a semicircle is 90°)
Check. The triangle ABC then has angles 34°, 90° and 56°, which add to 180°. ✓
Watch out. The right angle is at B because B is the point on the circle.
Method. Use the angle sum of triangle ABC.
- Use the angles we know. Angle BAC is given and angle ABC is 90° from part (a).Angle BAC = 34° and angle ABC = 90°
- Subtract from 180°. The three angles of a triangle add up to 180°.Angle ACB = 180° − 90° − 34° = 56° (the angles in a triangle add up to 180°)
Check. 34° + 90° + 56° = 180°. ✓
Watch out. Subtract both known angles, not just one.
Method. Angles in the same segment are equal. Find two angles made from the same chord.
- Find the chord. Angle ADB and angle ACB are both made from the chord AB. D and C are on the same side of AB.Both angles stand on chord AB.
- Write the equal angle. Angles in the same segment are equal.Angle ADB = angle ACB = 56° (angles in the same segment are equal)
Check. Angle ADB + angle BAC + ... is not needed; instead angle at the centre AOB = 2 × 56° = 112°, and arc AB matches 2 × 56°. ✓
Watch out. "Same segment" means the same chord and the same side of it.
Method. Angle BCD is made of two angles side by side at C.
- Split the angle. The diagonal CA splits angle BCD into angle BCA and angle ACD.Angle BCD = angle BCA + angle ACD
- Add. We know both parts.Angle BCD = 56° + 26° = 82°
Check. Opposite angles BAD and BCD should add to 180°; see part (e). ✓
Watch out. Add the parts that are side by side. Do not subtract.
Method. ABCD is a cyclic quadrilateral, so opposite angles add up to 180°.
- Spot the opposite corners. A and C are opposite corners, so angle BAD is opposite angle BCD.Angle BCD = 82° (part d)
- Subtract from 180°. Opposite angles of a cyclic quadrilateral add up to 180°.Angle BAD = 180° − 82° = 98° (opposite angles of a cyclic quadrilateral add up to 180°)
Check. Angle BAD = angle BAC + angle CAD = 34° + 64° = 98° (triangle ACD has a right angle at D by the semicircle theorem). ✓
Watch out. Check that the two opposite angles really are opposite.
Method. BAC is at the circumference and BOC is at the centre. Both stand on the arc BC.
- Name the arc. Both angles are made from the points B and C.arc BC
- Double it. The angle at the centre is twice the angle at the circumference.Angle BOC = 2 × 30° = 60° (the angle at the centre is twice the angle at the circumference)
Check. Half of 60° is 30°. ✓
Watch out. Use the circumference angle you are given and double it.
Method. OB and OC are radii, so triangle OBC is isosceles.
- Spot the isosceles triangle. Radii are equal, so the base angles are equal.angle OBC = angle OCB (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the angle at the centre from 180° and halve.Angle OBC = (180° − 60°) ÷ 2 = 60° (the angles in a triangle add up to 180°)
Check. 60° + 60° + 60° = 180°. ✓
Watch out. The apex of this triangle is at the centre O.
Method. Triangle ABC is isosceles because AB = AC.
- Write the base angles. The equal sides are AB and AC, so the base angles at B and C are equal.angle ABC = angle ACB (the base angles of an isosceles triangle are equal)
- Use the angle sum. The angle at A is 30°.Angle ABC = (180° − 30°) ÷ 2 = 75° (the angles in a triangle add up to 180°)
Check. 75° + 75° + 30° = 180°. ✓
Watch out. Here the apex is at A, on the circle, not at the centre.
Method. Angle ABC is made of angle ABO and angle OBC side by side.
- Write the sum. The radius BO lies inside angle ABC.angle ABO + angle OBC = angle ABC
- Subtract. Use angle ABC from part (c) and angle OBC from part (b).Angle ABO = 75° − 60° = 15°
Check. Triangle OAB is isosceles with OA = OB, so angle OAB = angle OBA = 15°, and AO bisects angle BAC = 30°. ✓
Watch out. Subtract the smaller part from the whole angle.
Method. BAC is at the circumference and BOC is at the centre. Both stand on the arc BC.
- Name the arc. Both angles are made from the points B and C.arc BC
- Double it. The angle at the centre is twice the angle at the circumference.Angle BOC = 2 × 50° = 100° (the angle at the centre is twice the angle at the circumference)
Check. Half of 100° is 50°. ✓
Watch out. Use the circumference angle you are given and double it.
Method. OB and OC are radii, so triangle OBC is isosceles.
- Spot the isosceles triangle. Radii are equal, so the base angles are equal.angle OBC = angle OCB (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the angle at the centre from 180° and halve.Angle OBC = (180° − 100°) ÷ 2 = 40° (the angles in a triangle add up to 180°)
Check. 40° + 40° + 100° = 180°. ✓
Watch out. The apex of this triangle is at the centre O.
Method. Triangle ABC is isosceles because AB = AC.
- Write the base angles. The equal sides are AB and AC, so the base angles at B and C are equal.angle ABC = angle ACB (the base angles of an isosceles triangle are equal)
- Use the angle sum. The angle at A is 50°.Angle ABC = (180° − 50°) ÷ 2 = 65° (the angles in a triangle add up to 180°)
Check. 65° + 65° + 50° = 180°. ✓
Watch out. Here the apex is at A, on the circle, not at the centre.
Method. Angle ABC is made of angle ABO and angle OBC side by side.
- Write the sum. The radius BO lies inside angle ABC.angle ABO + angle OBC = angle ABC
- Subtract. Use angle ABC from part (c) and angle OBC from part (b).Angle ABO = 65° − 40° = 25°
Check. Triangle OAB is isosceles with OA = OB, so angle OAB = angle OBA = 25°, and AO bisects angle BAC = 50°. ✓
Watch out. Subtract the smaller part from the whole angle.
Method. Use the tangent-radius right angles to find the angle at the centre, then halve it for the angle at the circumference.
- Find the angles at A and B. Each tangent is perpendicular to the radius at its point of contact.Angle OAT = angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
- Find the angle at the centre. Quadrilateral OATB has angles that add up to 360°.Angle AOB = 360° − 90° − 90° − 56° = 124° (the angles in a quadrilateral add up to 360°)
- Halve it. Angle ACB is at the circumference and stands on the same arc AB as angle AOB.Angle ACB = 124° ÷ 2 = 62° (the angle at the centre is twice the angle at the circumference)
Check. Second route (alternate segment): triangle TAB is isosceles, so angle TAB = (180° − 56°) ÷ 2 = 62°, and by the alternate segment theorem this equals angle ACB. ✓
Watch out. There are two circle theorems used here. Name each one as you use it.
Method. Use the tangent-radius right angles to find the angle at the centre, then halve it for the angle at the circumference.
- Find the angles at A and B. Each tangent is perpendicular to the radius at its point of contact.Angle OAT = angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
- Find the angle at the centre. Quadrilateral OATB has angles that add up to 360°.Angle AOB = 360° − 90° − 90° − 50° = 130° (the angles in a quadrilateral add up to 360°)
- Halve it. Angle ACB is at the circumference and stands on the same arc AB as angle AOB.Angle ACB = 130° ÷ 2 = 65° (the angle at the centre is twice the angle at the circumference)
Check. Second route (alternate segment): triangle TAB is isosceles, so angle TAB = (180° − 50°) ÷ 2 = 65°, and by the alternate segment theorem this equals angle ACB. ✓
Watch out. There are two circle theorems used here. Name each one as you use it.
Method. The quadrilateral OATB has two right angles, so the other two angles add up to 180°. Make an equation and solve it.
- Find the angles at A and B. Each tangent is perpendicular to the radius at its point of contact.Angle OAT = angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
- Use the quadrilateral OATB. Its angles add up to 360°. Two of them are 90°, so the other two add up to 360° − 180° = 180°.(5x + 10) + (8x + 14) = 180 (the angles in a quadrilateral add up to 360°)
- Collect like terms. Add the x terms together and the numbers together.13x + 24 = 180
- Subtract 24 from both sides. We want the number terms on the other side.13x = 180 − 24
13x = 156 - Divide both sides by 13. 13x means 13 × x, so dividing by 13 leaves x on its own.x = 156 ÷ 13 = 12
- Work out the angles. Put x back into each expression.Angle ATB = 5 × 12 + 10 = 70°
Angle AOB = 8 × 12 + 14 = 110°
Check. 70° + 110° = 180°. ✓
Watch out. The two tangent right angles use up 180° of the 360°. The equation is "the other two add to 180°".
Method. The quadrilateral OATB has two right angles, so the other two angles add up to 180°. Make an equation and solve it.
- Find the angles at A and B. Each tangent is perpendicular to the radius at its point of contact.Angle OAT = angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
- Use the quadrilateral OATB. Its angles add up to 360°. Two of them are 90°, so the other two add up to 360° − 180° = 180°.(5x + 10) + (8x + 14) = 180 (the angles in a quadrilateral add up to 360°)
- Collect like terms. Add the x terms together and the numbers together.13x + 24 = 180
- Subtract 24 from both sides. We want the number terms on the other side.13x = 180 − 24
13x = 156 - Divide both sides by 13. 13x means 13 × x, so dividing by 13 leaves x on its own.x = 156 ÷ 13 = 12
- Work out the angles. Put x back into each expression.Angle ATB = 5 × 12 + 10 = 70°
Angle AOB = 8 × 12 + 14 = 110°
Check. 70° + 110° = 180°. ✓
Watch out. The two tangent right angles use up 180° of the 360°. The equation is "the other two add to 180°".
Method. Opposite angles of a cyclic quadrilateral add up to 180°, so there are two separate equations: one with x and one with y.
- Pair up the opposite angles. A is opposite C, and B is opposite D. Each pair of opposite angles adds up to 180°, so we get two separate equations.(4x + 30) + (7x + 40) = 180
(6y + 18) + (3y + 54) = 180 (opposite angles of a cyclic quadrilateral add up to 180°) - Solve the first equation for x. Collect the x terms and numbers, then solve.11x + 70 = 180
- Subtract 70 from both sides. We want the number terms on the other side.11x = 180 − 70
11x = 110 - Divide both sides by 11. 11x means 11 × x, so dividing by 11 leaves x on its own.x = 110 ÷ 11 = 10
- Solve the second equation for y. It is the same idea with y.9y + 72 = 180
- Subtract 72 from both sides. We want the number terms on the other side.9y = 180 − 72
9y = 108 - Divide both sides by 9. 9y means 9 × y, so dividing by 9 leaves y on its own.y = 108 ÷ 9 = 12
Check. The angles are 70°, 90°, 110°, 90°. 70 + 110 = 180 and 90 + 90 = 180. ✓
Watch out. Do not mix the letters. The two equations are solved separately.
Method. Opposite angles of a cyclic quadrilateral add up to 180°, so there are two separate equations: one with x and one with y.
- Pair up the opposite angles. A is opposite C, and B is opposite D. Each pair of opposite angles adds up to 180°, so we get two separate equations.(4x + 30) + (7x + 40) = 180
(6y + 18) + (3y + 54) = 180 (opposite angles of a cyclic quadrilateral add up to 180°) - Solve the first equation for x. Collect the x terms and numbers, then solve.11x + 70 = 180
- Subtract 70 from both sides. We want the number terms on the other side.11x = 180 − 70
11x = 110 - Divide both sides by 11. 11x means 11 × x, so dividing by 11 leaves x on its own.x = 110 ÷ 11 = 10
- Solve the second equation for y. It is the same idea with y.9y + 72 = 180
- Subtract 72 from both sides. We want the number terms on the other side.9y = 180 − 72
9y = 108 - Divide both sides by 9. 9y means 9 × y, so dividing by 9 leaves y on its own.y = 108 ÷ 9 = 12
Check. The angles are 70°, 90°, 110°, 90°. 70 + 110 = 180 and 90 + 90 = 180. ✓
Watch out. Do not mix the letters. The two equations are solved separately.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(4x + 32) = 2 × (3x + 6) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (3x + 6) = 6x + 12
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.6x + 12 = 4x + 32 (written the other way round so that the x terms stay positive)
- Subtract 4x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.6x + 12 − 4x = 4x + 32 − 4x
2x + 12 = 32 - Subtract 12 from both sides. We want the number terms on the other side.2x = 32 − 12
2x = 20 - Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.x = 20 ÷ 2 = 10
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 3 × 10 + 6 = 36°
Check. Angle AOB = 4 × 10 + 32 = 72° and 2 × 36° = 72°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(4x + 32) = 2 × (3x + 6) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (3x + 6) = 6x + 12
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.6x + 12 = 4x + 32 (written the other way round so that the x terms stay positive)
- Subtract 4x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.6x + 12 − 4x = 4x + 32 − 4x
2x + 12 = 32 - Subtract 12 from both sides. We want the number terms on the other side.2x = 32 − 12
2x = 20 - Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.x = 20 ÷ 2 = 10
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 3 × 10 + 6 = 36°
Check. Angle AOB = 4 × 10 + 32 = 72° and 2 × 36° = 72°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Opposite angles of a cyclic quadrilateral add up to 180°.
- Spot the opposite corners. B is opposite D in the quadrilateral.Angle ADC = 95°
- Subtract from 180°. Opposite angles of a cyclic quadrilateral add up to 180°.Angle ABC = 180° − 95° = 85°
Check. 95° + 85° = 180°. ✓
Watch out. Opposite corners, not neighbours.
Method. AB and DC are parallel, so the angles BAD and ADC are co-interior angles (they sit between the parallel lines on the same side of the line AD).
- Spot the co-interior angles. Angle BAD and angle ADC are between the parallel lines AB and DC, on the same side of the transversal AD.Angle ADC = 95°
- Subtract from 180°. Co-interior angles between parallel lines add up to 180°.Angle BAD = 180° − 95° = 85° (co-interior angles between parallel lines add up to 180°)
Check. Opposite angle check: angle BCD = 180° − 85° = 95°, and BCD and ADC are equal as they should be in this isosceles trapezium. ✓
Watch out. Parallel lines give co-interior angles that add to 180°, not equal angles.
Method. Each pair of radii makes an isosceles triangle. Find the base angles at B in each triangle and add them.
- Spot two isosceles triangles. OA, OB and OC are all radii, so they are all equal. That makes triangles OAB and OBC isosceles.OA = OB = OC (radii)
- Find angle OBA. In triangle OAB the base angles are equal, and the angles add up to 180°.Angle OBA = (180° − 110°) ÷ 2 = 35° (the base angles of an isosceles triangle are equal; the angles in a triangle add up to 180°)
- Find angle OBC. Do the same in triangle OBC.Angle OBC = (180° − 90°) ÷ 2 = 45°
- Add the two parts of angle ABC. The line BO splits angle ABC into angle ABO and angle OBC.Angle ABC = 35° + 45° = 80°
Check. Second route: the arc AC not containing B has centre angle 360° − 110° − 90° = 160°. Half of it (the angle at the circumference) is 80°. ✓
Watch out. Angle ABC is made of two angles. Do not stop after finding only one of them.
Method. Angles made from the same chord, with their corners on the same side of it (in the same segment), are equal.
- Find the chord that the angle is drawn from. The angle at A has its arms going to B and C. So it is made from the chord BC.Angle BAC is made from chord BC.
- Find the other corner on the same side of that chord. The four points are A, B, C and D. Two of them are the ends of the chord (B and C). The other two are both on the same side of it, in the same segment.The corners left over are A and D.
- Write the equal angle. Angles in the same segment are equal, so use the other corner with the same two arms.Angle BAC = angle BDC (angles in the same segment are equal)
Check. Both angles go to the ends of chord BC. ✓
Watch out. Check that the two corners really are on the same side of the chord. For example, angle BAC and angle BDC are both made from chord BC.
Method. Angles made from the same chord, with their corners on the same side of it (in the same segment), are equal.
- Find the chord that the angle is drawn from. The angle at D has its arms going to A and B. So it is made from the chord AB.Angle ADB is made from chord AB.
- Find the other corner on the same side of that chord. The four points are A, B, C and D. Two of them are the ends of the chord (A and B). The other two are both on the same side of it, in the same segment.The corners left over are D and C.
- Write the equal angle. Angles in the same segment are equal, so use the other corner with the same two arms.Angle ADB = angle ACB (angles in the same segment are equal)
Check. Both angles go to the ends of chord AB. ✓
Watch out. Check that the two corners really are on the same side of the chord. For example, angle BAC and angle BDC are both made from chord BC.
Method. Angles made from the same chord, with their corners on the same side of it (in the same segment), are equal.
- Find the chord that the angle is drawn from. The angle at A has its arms going to C and D. So it is made from the chord CD.Angle CAD is made from chord CD.
- Find the other corner on the same side of that chord. The four points are A, B, C and D. Two of them are the ends of the chord (C and D). The other two are both on the same side of it, in the same segment.The corners left over are A and B.
- Write the equal angle. Angles in the same segment are equal, so use the other corner with the same two arms.Angle CAD = angle CBD (angles in the same segment are equal)
Check. Both angles go to the ends of chord CD. ✓
Watch out. Check that the two corners really are on the same side of the chord. For example, angle BAC and angle BDC are both made from chord BC.
Method. Angles made from the same chord, with their corners on the same side of it (in the same segment), are equal.
- Find the chord that the angle is drawn from. The angle at C has its arms going to A and D. So it is made from the chord AD.Angle ACD is made from chord AD.
- Find the other corner on the same side of that chord. The four points are A, B, C and D. Two of them are the ends of the chord (A and D). The other two are both on the same side of it, in the same segment.The corners left over are C and B.
- Write the equal angle. Angles in the same segment are equal, so use the other corner with the same two arms.Angle ACD = angle ABD (angles in the same segment are equal)
Check. Both angles go to the ends of chord AD. ✓
Watch out. Check that the two corners really are on the same side of the chord. For example, angle BAC and angle BDC are both made from chord BC.
Exercise 26K Reasoning and proof
Non-calculatorMethod. Look for the clue in the statement and match it to the theorem.
- Spot the clue. The clue is the word “diameter” with a point on the circle.diameter + a point on the circle
- Name the theorem. A diameter always makes a right angle at the circle.Angle in a semicircle is 90°.
Check. The right angle is at C, on the circle. ✓
Watch out. A diameter plus a point on the circle gives 90°.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. One angle is at the centre and the other at the circumference, on the same arc, and one is double the other.120° = 2 × 60°
- Name the theorem. The angle at the centre is twice the angle at the circumference.
Check. 120 ÷ 2 = 60. ✓
Watch out. The centre angle is the bigger one.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. Two angles are made from the same chord BC, with their corners on the same side.same chord, same side
- Name the theorem. Angles in the same segment are equal.
Check. Both angles go to the ends of the chord BC. ✓
Watch out. The same chord and the same side are both needed.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. Four points on a circle form a quadrilateral and the two angles are opposite corners.B and D are opposite
- Name the theorem. Opposite angles of a cyclic quadrilateral add up to 180°.
Check. B and D are not next to each other. ✓
Watch out. Only opposite angles add to 180°.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. A tangent and the radius to the point of contact meet at A.tangent + radius at the point of contact
- Name the theorem. The angle between a tangent and a radius is 90°.
Check. The angle is at the point of contact. ✓
Watch out. The radius must be drawn to the point of contact.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. Two tangents are drawn from the same point T outside the circle.two tangents from one point
- Name the theorem. Tangents from an external point are equal in length.
Check. The two tangents start from the same external point. ✓
Watch out. This is the reason for an isosceles triangle TAB.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. A tangent and a chord meet at the point of contact T, and an angle at B on the other side of the chord is equal to the tangent angle.tangent + chord, angle on the other side
- Name the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Check. B is on the other side of TA from the angle QTA. ✓
Watch out. This is the alternate segment theorem.
Method. Look for the clue in the statement and match it to the theorem.
- Spot the clue. A line from the centre meets a chord at a right angle.perpendicular from the centre to a chord
- Name the theorem. The perpendicular from the centre to a chord bisects the chord.
Check. M is the midpoint of the chord. ✓
Watch out. The line must come from the centre.
Method. Choose the word that makes the statement true.
- Think about the lengths. Lines from the centre to the circle are all the same length.OA and OB go from the centre O to the circle
- Choose the word. These lines are radii.radii
Check. Equal radii give an isosceles triangle. ✓
Watch out. This is the first reason in nearly every proof.
Method. Choose the word that makes the statement true.
- Think about the angle. A right angle on the circle comes from the angle in a semicircle theorem.angle in a semicircle
- Choose the word. The line must be a diameter.diameter
Check. AC must pass through the centre. ✓
Watch out. Do not forget that the line must be a diameter.
Method. Choose the word that makes the statement true.
- Think about the angle. A right angle between a line and the radius comes from the tangent theorem.tangent and radius
- Choose the word. tangent
Check. A tangent meets the radius at 90°. ✓
Watch out. The right angle is at the point of contact.
Method. Choose the words that make the statement true.
- Think about the position. The two angles are made from the same chord AD on the same side.same chord, same side
- Choose the words. same segment
Check. Angles in the same segment are equal. ✓
Watch out. Name the chord to be safe.
Method. Use the isosceles triangle ABD, then the same-segment theorem twice to carry the equal angles to C.
- Show triangle ABD is isosceles. AB = AD is given, so the base angles at B and D are equal.angle ABD = angle ADB (base angles of an isosceles triangle are equal)
- Match angle ABD with an angle at C. Angle ABD and angle ACD are both made from the chord AD, with B and C on the same side of it.angle ACD = angle ABD (angles in the same segment are equal)
- Match angle ADB with an angle at C. Angle ADB and angle ACB are both made from the chord AB, with D and C on the same side of it.angle ACB = angle ADB (angles in the same segment are equal)
- Compare. Both angles at C are equal to the equal base angles.angle ACD = angle ABD = angle ADB = angle ACB
- Conclude. The line CA splits angle BCD into two equal angles.So CA bisects angle BCD. ✓
Check. Test: arcs AB = AD = 90° give base angles 45°. Both angles at C are 45°. ✓
Watch out. Equal chords AB and AD make equal angles at the circumference; the proof shows why.
Method. Find the angle at the centre from the kite OATB, then halve it.
- Name the angle. Use a letter so the proof works for any size.Let angle ATB = t
- Use the kite OATB. The angles at A and B are 90° (tangent and radius). The angles of a quadrilateral add up to 360°.angle AOB = 360° − 90° − 90° − t = 180° − t
- Use the centre theorem. Angle ACB stands on the same arc AB as angle AOB, and the angle at the circumference is half the angle at the centre.angle ACB = (180° − t) ÷ 2 = 90° − ½t ✓
Check. Test: t = 64° gives angle ACB = 58°, as found in worked example 26.6. ✓
Watch out. It is the major arc that matters. If C were on the minor arc, the angle would be 90° + ½ t.
Method. Show that triangles ADC and BCD are congruent. They share the side DC.
- Use the cyclic quadrilateral. Opposite angles add up to 180°.angle ABC + angle ADC = 180°
- Use the parallel lines. AB and DC are parallel. Angle ABC and angle BCD are co-interior angles, so they add up to 180°.angle ABC + angle BCD = 180°
- Compare. Both angle ADC and angle BCD equal 180° − angle ABC.angle ADC = angle BCD
- Use the same segment. Angle DAC and angle DBC are made from the chord DC with A and B on the same side.angle DAC = angle DBC (angles in the same segment are equal)
- Show the triangles are congruent. In triangles ADC and BCD: angle ADC = angle BCD, angle DAC = angle CBD and the side DC is shared.The triangles are congruent (AAS).
- Match the sides. Triangle ADC matches triangle BCD with A matching B, D matching C and C matching D. Matching sides of congruent triangles are equal, so side AD matches side BC.So AD = BC. ✓
Check. Test: arcs 110°, 70°, 110°, 70° give an isosceles trapezium with AD = BC. ✓
Watch out. Name the two triangles in matching order so the equal sides can be read off.
Method. Do the problem twice, once with the alternate segment theorem and once with the tangent-radius right angle.
- Way 1: use the alternate segment theorem. The angle in the alternate segment equals the angle between the tangent and the chord.angle at the circumference on arc AB = x
- Way 1: use the centre theorem. The angle at the centre is twice the angle at the circumference.angle AOB = 2x
- Way 2: use the tangent and radius. The angle between a tangent and a radius is 90°.angle OAB = 90° − x
- Way 2: use the isosceles triangle OAB. OA = OB (radii), so the base angles are equal and the angle sum is 180°.angle AOB = 180° − 2(90° − x) = 180° − 180° + 2x = 2x
Check. Both ways give 2x. Test: x = 38° gives 76°. ✓
Watch out. When one route is blocked, try another. Two methods that agree give confidence.