Circles ICSE Class 10: Angle Properties and Cyclic Quadrilaterals

What Circles is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Circles is in the 3. Geometry (c) Circles (i) Angle Properties, (ii) Cyclic Properties part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Angle at the centre is double the angle at the circumference

If arc $AB$ subtends $\angle AOB$ at the centre $O$ and $\angle ACB$ at a point $C$ on the remaining part of the circle, then
$\angle AOB=2\angle ACB.$

Proof outline (to be taught): join $CO$ and extend it to $D$. $OA=OC$ (radii), so $\angle OAC=\angle OCA$; the exterior angle $\angle AOD=\angle OAC+\angle OCA=2\angle OCA$. Similarly $\angle BOD=2\angle OCB$. Adding, $\angle AOB=2\angle ACB$.

Careful: if $C$ is on the minor arc, the angle at the centre that goes with it is the reflex angle $AOB$.

Worked example

$\angle AOB=110^{\circ }$. Find $\angle ACB$ when $C$ is on the major arc, and $\angle ADB$ when $D$ is on the minor arc.

Try it first, then show the solutionSolution

1. $C$ on the major arc: $\angle ACB=\frac{110^{\circ }}{2}=55^{\circ }$.
2. $D$ on the minor arc: reflex $\angle AOB=250^{\circ }$, so $\angle ADB=125^{\circ }$.
3. Check: $ACBD$ is cyclic and $55^{\circ }+125^{\circ }=180^{\circ }$.

Answer: ∠ACB = 55°, ∠ADB = 125°

Angles in the same segment are equal

Angles subtended by the same chord (or arc) at points on the same side of the chord are equal (without proof). For example, if $C$ and $D$ are on the same arc, $\angle ACB=\angle ADB$.

Check carefully that both angles stand on the same chord and that both vertices are on the same side of it. Points on opposite sides give supplementary angles instead.

Worked example

$A$, $B$, $C$, $D$ lie on a circle, with $C$ and $D$ on the same side of $AB$. $\angle ACB=40^{\circ }$. Find $\angle ADB$.

Try it first, then show the solutionSolution

1. Both angles stand on chord $AB$ from the same segment.
2. $\angle ADB=\angle ACB=40^{\circ }$ (angles in the same segment).

Answer: 40°

Angle in a semicircle

The angle in a semicircle is a right angle: if $AB$ is a diameter and $C$ is any other point on the circle, $\angle ACB=90^{\circ }$.

Proof: the angle at the centre is the straight angle $\angle AOB=180^{\circ }$, so $\angle ACB=\frac{180^{\circ }}{2}=90^{\circ }$.

Whenever you see a diameter, look for the right angle at the circumference.

Worked example

$AB$ is a diameter and $C$ is on the circle with $\angle CAB=35^{\circ }$. Find $\angle ABC$.

Try it first, then show the solutionSolution

1. $\angle ACB=90^{\circ }$ (angle in a semicircle).
2. $\angle ABC=180^{\circ }-90^{\circ }-35^{\circ }=55^{\circ }$ (angle sum of a triangle).

Answer: 55°

Opposite angles of a cyclic quadrilateral

A cyclic quadrilateral has all four vertices on a circle. Its opposite angles are supplementary: $\angle A+\angle C=180^{\circ }$ and $\angle B+\angle D=180^{\circ }$.

Proof (to be taught): join $OB$ and $OD$. The two angles at the centre on arc $BAD$ and arc $BCD$ add up to $360^{\circ }$, and each is double the opposite angle of the quadrilateral, so $2\angle A+2\angle C=360^{\circ }$.

The converse is also used: if a pair of opposite angles of a quadrilateral is supplementary, it is cyclic.

Worked example

In cyclic quadrilateral $ABCD$, $\angle A=(3x+10{)}^{\circ }$ and $\angle C=(2x+20{)}^{\circ }$. Find both angles.

Try it first, then show the solutionSolution

1. $3x+10+2x+20=180$, so $5x=150$ and $x=30$.
2. $\angle A=100^{\circ }$, $\angle C=80^{\circ }$.

Answer: ∠A = 100°, ∠C = 80°

Exterior angle of a cyclic quadrilateral

If one side of a cyclic quadrilateral is produced, the exterior angle equals the interior opposite angle (without proof). If $AB$ is produced to $E$, then $\angle CBE=\angle ADC$.

This follows from the supplementary property: $\angle CBE=180^{\circ }-\angle ABC=\angle ADC$.

Worked example

$ABCD$ is cyclic and $AB$ is produced to $E$. $\angle CBE=105^{\circ }$. Find $\angle ADC$ and $\angle ABC$.

Try it first, then show the solutionSolution

1. $\angle ADC=\angle CBE=105^{\circ }$ (exterior angle of a cyclic quadrilateral).
2. $\angle ABC=180^{\circ }-105^{\circ }=75^{\circ }$ (linear pair).

Answer: ∠ADC = 105°, ∠ABC = 75°

Radii make isosceles triangles

Any two radii, such as $OA$ and $OB$, are equal, so triangle $OAB$ is isosceles with $\angle OAB=\angle OBA$. This is often the missing step between the given angle and a circle theorem.

Useful pattern: if $\angle OAB=x$, then $\angle AOB=180^{\circ }-2x$, and the angle at the circumference on the major arc is $90^{\circ }-x$.

Worked example

$O$ is the centre and $\angle OAB=28^{\circ }$. $C$ is on the major arc $AB$. Find $\angle AOB$ and $\angle ACB$.

Try it first, then show the solutionSolution

1. $OA=OB$ (radii), so $\angle OBA=28^{\circ }$.
2. $\angle AOB=180^{\circ }-56^{\circ }=124^{\circ }$.
3. $\angle ACB=\frac{124^{\circ }}{2}=62^{\circ }$ (angle at centre is double).

Answer: ∠AOB = 124°, ∠ACB = 62°

Writing reasons the CISCE way

For each angle you find, write the reason in brackets. Standard phrases:

• (angle at the centre is double the angle at the circumference)
• (angles in the same segment)
• (angle in a semicircle)
• (opposite angles of a cyclic quadrilateral)
• (exterior angle of a cyclic quadrilateral)
• (radii; base angles of an isosceles triangle)
• (angle sum of a triangle; linear pair; alternate angles)

Name angles with three letters, such as $\angle ACB$, when several angles share a vertex.

Worked example

$ABCD$ is cyclic, $\angle BAC=30^{\circ }$ and $\angle BCA=50^{\circ }$. Find $\angle ADC$ with reasons.

Try it first, then show the solutionSolution

1. $\angle ABC=180^{\circ }-30^{\circ }-50^{\circ }=100^{\circ }$ (angle sum of $△ABC$).
2. $\angle ADC=180^{\circ }-100^{\circ }=80^{\circ }$ (opposite angles of a cyclic quadrilateral).

Answer: 80°

Maths in real life: Circles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Angle at the Centre Theorem
• Cyclic Quadrilateral Properties
• Angle in a Semicircle and Angles in the Same Segment

Circles formula sheet

Angle at the centre

$\angle AOB=2\angle ACB$

C on the remaining (major) arc; use the reflex angle for a point on the minor arc.

Same segment

$\angle ACB=\angle ADB$

C and D on the same side of chord AB.

Angle in a semicircle

$AB\text{ diameter}\Rightarrow \angle ACB=90^{\circ }$

Look for a diameter.

Cyclic quadrilateral

$\angle A+\angle C=180^{\circ },\angle B+\angle D=180^{\circ }$

Opposite angles are supplementary, not equal.

Exterior angle of a cyclic quadrilateral

$\angle CBE=\angle ADC$

AB produced to E.

Isosceles triangle on two radii

$\angle AOB=180^{\circ }-2\angle OAB$

OA = OB.

How the board asks Circles

The question patterns to expect, and the method that works for each.

Tricks to remember Circles

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Centre is twice

  The centre is closer, so it sees the arc at twice the angle.

  Use it when: Angle at the centre theorem.

• Same side, same angle; opposite sides, add to 180

  Two points on the same arc see a chord at equal angles; points on opposite arcs see it at supplementary angles.

  Use it when: Choosing between 'same segment' and 'cyclic quadrilateral'.

• Diameter means 90

  See a diameter? Draw the right angle at the circumference.

  Use it when: Semicircle questions.

• Two radii, one isosceles

  Every time two radii appear in a triangle, mark the equal base angles.

  Use it when: The missing link in multi-step questions.

Circles: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Circles

The slips examiners see most often, and what to write instead.

• ✗Mistake: Doubling the wrong way

  CISCE examiners' comments, 2024: candidates misapplied 'angle at the centre is double the angle at the circumference', for example taking ∠ABC as 130° (the angle at the centre) instead of half of it.

  Wrong working

  ∠AOC = 130°, so ∠ABC = 130° (or 260°)

  Right working

  ∠ABC = ½ × 130° = 65°

  ✓Do this: The angle at the centre is the bigger one. Halve when going to the circumference; double when going to the centre.

  Usually costs: 1 mark, and every angle that depends on it.

• ✗Mistake: Opposite angles taken as equal

  CISCE examiners' comments, 2025: in an MCQ, candidates believed opposite angles of a cyclic quadrilateral are equal rather than supplementary.

  ✓Do this: Cyclic: opposite angles add to 180°. Parallelogram: opposite angles are equal.

  Usually costs: 1 mark.

• ✗Mistake: No reasons given

  CISCE examiners' comments, 2024 and 2025: candidates did not give reasons for the angles they found.

  ✓Do this: Write the theorem name in brackets after every angle.

  Usually costs: Marks in every part of a reasons question.

• ✗Mistake: Wrongly assuming 'same segment'

  CISCE examiners' comments, 2025: candidates wrote ∠BOX = ∠CYX, assuming they were angles in the same segment when one angle was at the centre.

  ✓Do this: Same segment needs both vertices on the circumference, on the same side of the same chord.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Missing the isosceles triangle

  CISCE examiners' comments, 2024 and 2025: candidates did not use the equal base angles of an isosceles triangle (from equal radii or equal chords).

  ✓Do this: Mark equal radii on the figure at the start.

  Usually costs: 1 mark per missed step.

Practice questions on Circles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   $AB$ is a diameter of a circle and $C$ is any other point on it. $\angle ACB$ is:
   (a) $45^{\circ }$
   (b) $60^{\circ }$
   (c) $90^{\circ }$
   (d) $180^{\circ }$

   Hint

   Semicircle.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   In cyclic quadrilateral $PQRS$, $\angle P=85^{\circ }$. Then $\angle R$ is:
   (a) $85^{\circ }$
   (b) $95^{\circ }$
   (c) $105^{\circ }$
   (d) $170^{\circ }$

   Hint

   Supplementary.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   $O$ is the centre and $\angle OAB=40^{\circ }$. For $C$ on the major arc $AB$, $\angle ACB$ is:
   (a) $40^{\circ }$
   (b) $50^{\circ }$
   (c) $80^{\circ }$
   (d) $100^{\circ }$

   Hint

   Find ∠AOB first.

   Answer

   (b): $\angle AOB=100^{\circ }$.

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): A parallelogram inscribed in a circle must be a rectangle.
   Reason (R): Opposite angles of a parallelogram are equal, and opposite angles of a cyclic quadrilateral are supplementary.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Equal and supplementary means 90° each.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   $ABCD$ is a cyclic quadrilateral with $AB=AD$ and $\angle BCD=110^{\circ }$. Find $\angle BAD$, $\angle ABD$ and $\angle ACD$, giving reasons.

   Hint

   △ABD is isosceles.

   Answer

   $\angle BAD=70^{\circ }$ (cyclic quadrilateral). $\angle ABD=\angle ADB=55^{\circ }$ (isosceles, $AB=AD$). $\angle ACD=\angle ABD=55^{\circ }$ (same segment, chord $AD$).

6. Q6 · 4 marks · Short answerHarder board level

   $AB$ is a diameter of a circle with centre $O$, and $C$ is a point on the circle with $\angle ABC=3\angle BAC$. Find $\angle BAC$, $\angle ABC$ and $\angle AOC$.

   Hint

   The two angles add to 90°.

   Answer

   $4\angle BAC=90^{\circ }$: $\angle BAC=22.5^{\circ }$, $\angle ABC=67.5^{\circ }$. $\angle AOC=2\angle ABC=135^{\circ }$.

7. Q7 · 3 marks · Long answerBoard level

   $O$ is the centre and $\angle AOB=100^{\circ }$. $C$ is on the major arc and $D$ on the minor arc. Find $\angle ACB$ and $\angle ADB$.

   Hint

   Use the reflex angle for D.

   Answer

   $\angle ACB=50^{\circ }$; $\angle ADB=\frac{260^{\circ }}{2}=130^{\circ }$.

8. Q8 · 4 marks · Long answerHarder board level

   $ABCD$ is cyclic and $AB$ is produced to $E$. $\angle CBE=3x^{\circ }$ and $\angle ADC=(2x+20{)}^{\circ }$. (a) Find $x$. (b) Find $\angle ABC$. (c) If also $AD=DC$, show that triangle $ADC$ is equilateral.

   Hint

   Exterior angle = interior opposite angle.

   Answer

   (a) $3x=2x+20$, $x=20$. (b) $\angle CBE=60^{\circ }$, so $\angle ABC=120^{\circ }$. (c) $\angle ADC=60^{\circ }$; $AD=DC$ gives $\angle DAC=\angle DCA=60^{\circ }$: equilateral.

9. Q9 · 4 marks · Long answerHOTS / competency

   Prove that the angle in a semicircle is a right angle. Hence, if $AB$ is a diameter, $C$ is on the circle and $CD\perp AB$, prove that $\angle ACD=\angle ABC$.

   Hint

   Both are 90° minus ∠CAB.

   Answer

   $\angle AOB=180^{\circ }$ is the angle at the centre on arc $AB$, so $\angle ACB=90^{\circ }$. In $△ACD$, $\angle ACD=90^{\circ }-\angle CAD$. In $△ACB$, $\angle ABC=90^{\circ }-\angle CAB$. Since $\angle CAD=\angle CAB$, $\angle ACD=\angle ABC$.

10. Q10 · 4 marks · Long answerHarder board level

    $A$, $B$, $C$ lie on a circle with centre $O$ inside triangle $ABC$. $\angle BAC=50^{\circ }$ and $\angle ABC=60^{\circ }$. Find $\angle BOC$, $\angle OBC$, $\angle AOB$ and $\angle OBA$.

    Hint

    ∠ACB = 70°.

    Answer

    $\angle BOC=100^{\circ }$, $\angle OBC=40^{\circ }$, $\angle AOB=2\times 70^{\circ }=140^{\circ }$, $\angle OBA=20^{\circ }$. Check: $40^{\circ }+20^{\circ }=60^{\circ }$ ✓

11. Q11 · 1 mark · MCQHOTS / competency

    In cyclic quadrilateral $ABCD$, $\angle A-\angle C=40^{\circ }$. Then $\angle A$ is:
    (a) $70^{\circ }$
    (b) $100^{\circ }$
    (c) $110^{\circ }$
    (d) $140^{\circ }$

    Hint

    Sum is 180°, difference 40°.

    Answer

    (c)

12. Q12 · 5 marks · Long answerHOTS / competency

    Two circles meet at $P$ and $Q$. A straight line through $P$ meets the first circle at $A$ and the second at $B$ (with $P$ between $A$ and $B$). A straight line through $Q$ meets the first circle at $C$ and the second at $D$ (with $Q$ between $C$ and $D$). Prove that $AC\parallel BD$.

    Hint

    Use cyclic quadrilaterals APQC and PBDQ.

    Answer

    In the first circle, $APQC$ is cyclic: $\angle ACQ+\angle APQ=180^{\circ }$. Also $\angle APQ+\angle QPB=180^{\circ }$ (linear pair), so $\angle ACQ=\angle QPB$. In the second circle, $PBDQ$ is cyclic: $\angle QPB+\angle QDB=180^{\circ }$. Hence $\angle ACQ+\angle QDB=180^{\circ }$. These are co-interior angles on the transversal $CD$, so $AC\parallel BD$.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Angle at the centre vs at the circumference on the same arc?Show answer

   The angle at the centre is double.

2. 2.Angles in the same segment?Show answer

   Equal.

3. 3.Angle in a semicircle?Show answer

   90°

4. 4.Opposite angles of a cyclic quadrilateral?Show answer

   Supplementary (sum 180°).

5. 5.Exterior angle of a cyclic quadrilateral?Show answer

   Equal to the interior opposite angle.

6. 6.Which two theorems are 'without proof'?Show answer

   Angles in the same segment; exterior angle of a cyclic quadrilateral.

7. 7.∠OAB = 30° with OA, OB radii. ∠AOB?Show answer

   120°

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you prove the angle-at-the-centre theorem?
• ☐Do you use the reflex angle when the point is on the minor arc?
• ☐Do you check 'same side of the chord' before using same segment?
• ☐Do you remember opposite angles of a cyclic quadrilateral are supplementary, not equal?
• ☐Do you write a reason for every angle?
• ☐Do you name angles with three letters?

Circles: frequently asked questions