Cyclic Quadrilateral Properties: Opposite Angles and the Exterior Angle (ICSE Class 10)

Understand Cyclic Quadrilateral Properties

Cyclic Quadrilateral Properties: formulas to remember

Opposite angles

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

Proof to be taught.

Exterior angle

$\angle CBE=\angle ADC$

$AB$ produced to $E$; without proof.

Angle from the centre

$\angle BAD=\frac{1}{2}\angle BOD,\angle BCD=180^{\circ }-\frac{1}{2}\angle BOD$

$A$ on the major arc $BD$, $C$ on the minor arc.

How to solve Cyclic Quadrilateral Properties questions

1. 1Read the vertices in order round the circle and mark the opposite pairs: $A$ with $C$, $B$ with $D$.
2. 2If an angle is given, its opposite angle is $180^{\circ }$ minus it.
3. 3If a side is produced, the exterior angle equals the interior angle at the opposite vertex.
4. 4For angles formed by a diagonal, use angles in the same segment or the angle sum of a triangle.
5. 5If angles are given in $x$, set opposite angles to add to $180^{\circ }$ and solve; then check all four add to $360^{\circ }$.
6. 6Write the reason for every angle.

Cyclic Quadrilateral Properties in real life

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.

Tricks to remember Cyclic Quadrilateral Properties

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

• Across adds to 180

  Corners across the circle add to $180^{\circ }$; corners next to each other need not.

  Use it when: Avoiding the 'opposite angles are equal' trap.

• Outside equals opposite

  Produce a side: the outside angle copies the inside angle at the far corner.

  Use it when: Exterior-angle questions in one line.

• Cyclic special shapes

  Cyclic parallelogram → rectangle; cyclic trapezium → isosceles; cyclic kite → right angles at the side corners.

  Use it when: Assertion–reason and proof questions.

Cyclic Quadrilateral Properties: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Where students lose marks in Cyclic Quadrilateral Properties

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

• ✗Mistake: Taking opposite angles as equal

  In a cyclic quadrilateral opposite angles add to $180^{\circ }$; they are equal only when both are $90^{\circ }$.

  Wrong working

  $\angle A=60^{\circ }$, so $\angle C=60^{\circ }$.

  Right working

  $\angle C=180^{\circ }-60^{\circ }=120^{\circ }$.

  ✓Do this: Say 'supplementary' every time you use the property.

  Usually costs: Reported in 2025 Q1(vi): candidates believed the opposite angles were equal and chose $60^\circ$.

• ✗Mistake: Pairing adjacent angles

  $\angle A$ and $\angle B$ are next to each other; only $A$–$C$ and $B$–$D$ are opposite.

  Wrong working

  $\angle A=70^{\circ }$, so $\angle B=110^{\circ }$.

  Right working

  $\angle C=110^{\circ }$; $\angle B$ needs other information.

  ✓Do this: Trace the vertices round the circle in order before pairing.

  Usually costs: The part, and any later angle that depends on it.

• ✗Mistake: Exterior angle matched to the wrong vertex

  The exterior angle at $B$ equals the interior angle at $D$, the vertex opposite $B$.

  Wrong working

  $AB$ produced to $E$: $\angle CBE=\angle BAD$.

  Right working

  $\angle CBE=\angle ADC$.

  ✓Do this: The exterior angle copies the angle at the far corner.

  Usually costs: 1 mark per angle in a 3-mark part.

Practice questions on Cyclic Quadrilateral Properties

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

   $ABCD$ is a cyclic quadrilateral and $BC$ is produced to $E$. If $\angle DCE=74^{\circ }$, then $\angle BAD$ is:
   (a) $16^{\circ }$
   (b) $74^{\circ }$
   (c) $106^{\circ }$
   (d) $148^{\circ }$

   Hint

   Exterior angle = interior opposite angle.

   Answer

   (b)

2. Q2 · 2 marks · Short answerBoard level

   In cyclic quadrilateral $ABCD$, $\angle A=(5x-8{)}^{\circ }$ and $\angle C=(3x+12{)}^{\circ }$. Find $x$, $\angle A$ and $\angle C$.

   Hint

   $\angle A+\angle C=180^{\circ }$.

   Answer

   $8x+4=180$, $x=22$; $\angle A=102^{\circ }$, $\angle C=78^{\circ }$.

3. Q3 · 2 marks · Short answerHarder board level

   $O$ is the centre of a circle and $\angle BOD=150^{\circ }$. $A$ is on the major arc $BD$ and $C$ on the minor arc $BD$. Find $\angle BAD$ and $\angle BCD$.

   Hint

   Angle at the centre first, then the cyclic quadrilateral $ABCD$.

   Answer

   $\angle BAD=75^{\circ }$ (angle at centre); $\angle BCD=105^{\circ }$ (opposite angles of cyclic quadrilateral $ABCD$).

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): In cyclic quadrilateral $ABCD$, $\angle A=70^{\circ }$ and $\angle B=95^{\circ }$, so $\angle D=110^{\circ }$.
   Reason (R): The 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

   Which angle is opposite $\angle D$?

   Answer

   (b): $\angle D$ is opposite $\angle B$, so $\angle D=85^{\circ }$.

5. Q5 · 3 marks · Short answerHarder board level

   $ABCD$ is a cyclic quadrilateral with diagonals $AC$ and $BD$. $\angle BAC=42^{\circ }$ and $\angle CAD=33^{\circ }$. Find $\angle BCD$, $\angle BDC$ and $\angle CBD$, and check that the angles of $△BCD$ add to $180^{\circ }$.

   Hint

   $\angle BAD=42^{\circ }+33^{\circ }$.

   Answer

   $\angle BCD=180^{\circ }-75^{\circ }=105^{\circ }$; $\angle BDC=\angle BAC=42^{\circ }$ and $\angle CBD=\angle CAD=33^{\circ }$ (same segment). $105^{\circ }+42^{\circ }+33^{\circ }=180^{\circ }$ ✓

6. Q6 · 4 marks · Long answerHarder board level

   $PQRS$ is a cyclic quadrilateral and $PQ$ is produced to $T$. $\angle RQT=68^{\circ }$ and $\angle PRS=34^{\circ }$. Find $\angle PSR$, $\angle PQR$, $\angle RPS$ and $\angle RQS$.

   Hint

   Start with the exterior angle at $Q$.

   Answer

   $\angle PSR=68^{\circ }$ (exterior angle); $\angle PQR=112^{\circ }$; $\angle RPS=180^{\circ }-68^{\circ }-34^{\circ }=78^{\circ }$; $\angle RQS=\angle RPS=78^{\circ }$ (same segment, chord $RS$).

7. Q7 · 4 marks · Long answerHOTS / competency

   $ABCD$ is a cyclic quadrilateral. $BA$ and $CD$ are produced to meet at $P$ (beyond $A$ and $D$).
   (a) Prove that $△PAD\sim △PCB$.
   (b) If $PA=4$ cm, $AB=5$ cm and $PD=3$ cm, find $PC$ and $DC$.

   Hint

   $\angle PAD$ is an exterior angle of $ABCD$.

   Answer

   (a) $\angle PAD=\angle BCD$ (exterior angle of a cyclic quadrilateral) and $\angle P$ is common, so the triangles are similar (AA). (b) $\frac{PA}{PC}=\frac{PD}{PB}$: $PC=\frac{4\times 9}{3}=12$ cm, so $DC=9$ cm.

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.Opposite angles of a cyclic quadrilateral?Show answer

   Supplementary: they add to $180^{\circ }$.

2. 2.$\angle A=64^{\circ }$ in cyclic $ABCD$. $\angle C$?Show answer

   $116^{\circ }$

3. 3.Exterior angle of a cyclic quadrilateral equals…?Show answer

   The interior opposite angle.

4. 4.A cyclic parallelogram is always a…?Show answer

   Rectangle.

5. 5.A cyclic trapezium is always…?Show answer

   Isosceles (non-parallel sides equal).

Cyclic Quadrilateral Properties: frequently asked questions