Circle theorem proof Quiz (Grade 8+)

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Circle theorem proof — Notes & Worked Examples Show / Hide

Explanation:

Higher tier papers can ask you to prove standard circle theorems from first principles. The key construction is to draw radii from the centre O to the vertices on the circumference, creating isosceles triangles (since all radii are equal).

Label the equal base angles of the isosceles triangles as x and y, then use angles in a triangle (180°) and angles around a point (360°) or tangent-radius (90°) to link the angles algebraically. Always state geometric reasons at every step.

Example 1: Prove that opposite angles in a cyclic quadrilateral ABCD add up to 180°.

Solution:

Join B, D to O. Let ∠BAD = x, ∠BCD = y ⇒ ∠BOD = 2x and 2y

Angles around O: 2x + 2y = 360° ⇒ x + y = 180°

Example 2: Prove the alternate segment theorem: for tangent PT at A and chord AB, ∠BAT = ∠ACB.

Solution:

Let ∠BAT = x. ∠OAT = 90° ⇒ ∠OAB = ∠OBA = 90° − x

∠AOB = 180° − 2(90° − x) = 2x ⇒ ∠ACB = 12(2x) = x

One Page Guide (PDF) Circle theorem proof Worksheets

Question 1

[1 mark]

Points A and B lie on a circle with centre O. In isosceles triangle OAB, OA = OB (radii) and angle OAB = 30°.
Prove by calculation that angle AOB = 120°.

Worked Solution & Answer

OA = OB (radii), so ∠ OBA = ∠ OAB = 30°
Angles in △ OAB sum to 180° ⇒ ∠ AOB = 120°. B1

Answer: ∠ OBA = 30° (base angles of isosceles △ OAB), ∠ AOB = 180° - 2(30°) = 120°

Question 2

[1 mark]

Points A and B lie on a circle with centre O. In isosceles triangle OAB, OA = OB (radii) and angle OAB = 32°.
Prove by calculation that angle AOB = 116°.

Worked Solution & Answer

OA = OB (radii), so ∠ OBA = ∠ OAB = 32°
Angles in △ OAB sum to 180° ⇒ ∠ AOB = 116°. B1

Answer: ∠ OBA = 32° (base angles of isosceles △ OAB), ∠ AOB = 180° - 2(32°) = 116°

Question 3 AQA Nov 2022 1H Q26

[4 marks]

P, Q and R are points on a circle.
SP is a tangent to the circle.
RQ = PQ

Prove that y = 90° - x

Worked Solution & Answer

By the alternate segment theorem, angle PQR = angle RPS = 2x M2
Since RQ = PQ, triangle PQR is isosceles, so angle QPR = angle PRQ = y M1
Angles in triangle PQR add to 180°: y + y + 2x = 180° ⇒ 2y = 180° – 2x ⇒ y = 90° – x. C1

Answer: Angle PQR = 2x (alternate segment theorem).
Angle QPR = y (isosceles triangle PQR as RQ = PQ).
2y + 2x = 180° (angles in a triangle sum to 180°), so y = 90° – x.

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Question 4

[4 marks]

A, B and C are points on the circumference of a circle, centre O. AOB is a diameter. Prove that angle ACB is 90° without using circle theorems.

A, B and C are points on the circumference of a circle, centre O.
AOB is a diameter of the circle.
Prove that angle ACB is 90°
You must **not** use any circle theorems in your proof.

Worked Solution & Answer

(The circumference of a circle is π × diameter = 2πr)
Join O to C M1
OA = OC = OB (radii of circle), so triangles OAC and OBC are isosceles M1
Let ∠OAC = ∠OCA = x and ∠OBC = ∠OCB = y (base angles of isosceles triangles are equal) M1
Sum of angles in triangle ABC = x + (x + y) + y = 2x + 2y = 180° ⇒ x + y = 90°, so ∠ACB = 90°. C1

Answer: Draw radius OC. OA = OC = OB (radii), so △ OAC and △ OBC are isosceles. Let ∠ OAC = ∠ OCA = x and ∠ OBC = ∠ OCB = y (base angles of an isosceles triangle are equal). In △ ABC, x + (x + y) + y = 180° (angles in a triangle sum to 180°) ⇒ 2(x + y) = 180° ⇒ ∠ ACB = x + y = 90°.

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Question 5

[4 marks]

A and B are points on a circle, centre O.
MAP and NBP are tangents to the circle.

Prove that AP = BP

Worked Solution & Answer

Join OA, OB and OP. ∠OAP = ∠OBP = 90° (angle between radius and tangent is 90°) M1
OA = OB (both radii) M1
OP is a common side M1
By RHS, ΔOAP ≅ ΔOBP, hence AP = BP. C1

Answer: In triangles OAP and OBP: ∠OAP = ∠OBP = 90° (angle between tangent and radius is 90°), OA = OB (radii of the same circle), and OP is common. Therefore triangles OAP and OBP are congruent by RHS, so AP = BP.

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Question 6 IGCSE Nov 2020 1HR Q15

[4 marks]

B, D, E and F are points on a circle.
ABC is the tangent to the circle at B.
Angle EDF = 40°
Angle FBC = 70°

Prove that the tangent ABC is parallel to EF.
Give a reason for each stage of your working.

Worked Solution & Answer

Angle BDF = 70° (alternate segment theorem) => Angle EDB = 40 + 70 = 110° => Angle EFB = 180 - 110 = 70° (cyclic quad) => alternate angles FBC and EFB are both 70°, so EF || ABC. M3 C1

Answer: Angle BDF = 70° (alternate segment theorem); Angle EFB = 180° – (70° + 40°) = 70° (opposite angles in a cyclic quadrilateral sum to 180°); Angle FBC = Angle EFB = 70°, since alternate angles are equal, ABC is parallel to EF.

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Question 7 Practice Set 1 1H Q23

[4 marks]

Work out
A and D are two points on the circumference of a circle.
A and B are two points on the circumference of a smaller circle.
DB and AC are tangents to both circles.
E is the intersection of DB and AC.
E is the midpoint of AC.
Prove that ABCD is a rectangle.

A and D are two points on the circumference of a circle.
A and B are two points on the circumference of a smaller circle.
DB and AC are tangents to both circles.
E is the intersection of DB and AC.
E is the midpoint of AC.
Prove that ABCD is a rectangle.

Worked Solution & Answer

(The circumference of a circle is π × diameter = 2πr)
Tangents from E to big circle: EA = ED M1
Tangents from E to small circle: EA = EB M1
Given EA = EC M1
Since all four half-diagonals are equal, ABCD is a rectangle. C1

Answer: EA = ED and EA = EB (tangents from an external point are equal), and EA = EC (E is midpoint of AC). Thus EA = EB = EC = ED, so the diagonals AC and BD bisect each other and are equal in length, hence ABCD is a rectangle.

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Question 8 AQA Nov 2018 1H Q20

[4 marks]

P, Q, R and S are points on a circle.
PXR and QXS are straight lines.
PX = SX

Prove that QS is not a diameter of the circle.

Worked Solution & Answer

1. In triangle PXS, PX = SX, so triangle PXS is isosceles M1
Therefore, the base angles are equal:
∠ XPS = ∠ XSP = 180° - 50°2 = 65°
2. Using the circle theorem "angles in the same segment subtended by the same arc are equal": M1
Arc QR subtends ∠ QPR and ∠ QSR at the circumference
From the diagram, ∠ XSR = 27°, so ∠ QPR = 27°.
3. Find ∠ QPS: M1
∠ QPS = ∠ QPR + ∠ XPS = 27° + 65° = 92°
4. By the circle theorem "the angle subtended by a diameter at the circumference is 90°" (angle in a semicircle):
If QS were a diameter, ∠ QPS would equal 90°
Since ∠ QPS = 92° ≠ 90°, QS is not a diameter of the circle. C1

Answer: Proof that angle QPS = 92° ≠ 90°

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Question 9 AQA Nov 2019 2H Q25

[4 marks]

C, D and E are points on a circle.
CE = DE
The tangent at D is shown.
ACD and BCE are straight lines.

Prove that y = 3x

Worked Solution & Answer

1. By the alternate segment theorem, the angle between the tangent at D and chord CD equals the angle in the alternate segment: ∠ CED = 4x. M1
2. Triangle CDE is isosceles with CE = DE, so ∠ ECD = ∠ EDC = 180° - 4x2 = 90° - 2x. M1
3. Since ACD and BCE are straight lines intersecting at C, vertically opposite angles are equal: ∠ ACB = ∠ ECD = 90° - 2x.
4. In triangle ABC, angles sum to 180°: M1
y + (90° - x) + (90° - 2x) = 180°
y + 180° - 3x = 180° ⇒ y = 3x. C1

Answer: y = 3x

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Question 10 IGCSE Jan 2019 2H Q20

[5 marks]

A, B, C and D are points on a circle.
PCQ is a tangent to the circle.
AB = CB.

Angle BCQ = x°
Prove that angle CDA = 2x°
Give reasons for each stage in your working.

Worked Solution & Answer

∠CAB = x° (alternate segment theorem). ∠ACB = x° (base angles of isosceles ΔABC are equal). ∠ABC = 180° − 2x° (angles in a triangle sum to 180°). ∠CDA = 180° − (180° − 2x°) = 2x° (opposite angles of a cyclic quadrilateral sum to 180°). M4 C1

Answer: By the alternate segment theorem, ∠ CAB = x°. Since AB = CB, △ ABC is isosceles so ∠ ACB = x° and ∠ ABC = 180° - 2x°. Opposite angles of cyclic quadrilateral ABCD sum to 180°, so ∠ CDA = 180° - (180° - 2x°) = 2x°.

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Score: 0 / 10 (0 / 35 marks)

Circle theorem proof Worksheets Next: Vector proof →

Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum