Cyclic Quadrilateral
A cyclic quadrilateral is a quadrilateral whose all four vertices lie on a single circle. This circle is called the circumscribed circle or circumcircle of the quadrilateral.
Cyclic quadrilaterals have a remarkable property: the sum of each pair of opposite angles is exactly 180° (supplementary). This is one of the most important and frequently tested results in the NCERT Class 9 chapter on Circles (Chapter 10).
The concept of cyclic quadrilaterals connects circle geometry with angle properties. It is used extensively in geometric proofs, construction problems, and competitive mathematics (olympiads, JEE, NTSE). Understanding this concept is essential for mastering circle-related theorems.
The four points lying on a circle are called concyclic points. Not every quadrilateral is cyclic — only those whose opposite angles are supplementary qualify. This gives us both the theorem (if cyclic, then opposite angles are supplementary) and its converse (if opposite angles are supplementary, then cyclic).
The study of cyclic quadrilaterals dates back to Ptolemy (2nd century AD), who proved the famous Ptolemy's Theorem relating the sides and diagonals. The Indian mathematician Brahmagupta (7th century AD) gave a formula for the area of a cyclic quadrilateral that generalises Heron's Formula.
What is Cyclic Quadrilateral?
Definition: A cyclic quadrilateral is a quadrilateral inscribed in a circle, meaning all four vertices lie on the circumference of the circle.
The main property:
∠A + ∠C = 180° and ∠B + ∠D = 180°
That is, the sum of opposite angles in a cyclic quadrilateral is 180°.
Key terms:
- Circumscribed circle (circumcircle): The circle passing through all four vertices.
- Circumcentre: The centre of the circumcircle.
- Circumradius: The radius of the circumcircle.
- Concyclic points: Points that lie on the same circle.
Important:
- Every rectangle and square is a cyclic quadrilateral (all angles are 90°, and 90° + 90° = 180°).
- A parallelogram is cyclic only if it is a rectangle.
- Not every quadrilateral is cyclic. A quadrilateral is cyclic if and only if its opposite angles are supplementary.
Cyclic Quadrilateral Formula
Key Results and Formulas:
1. Opposite angles property:
∠A + ∠C = 180° and ∠B + ∠D = 180°
2. Exterior angle property:
- An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
- If side AB of cyclic quadrilateral ABCD is extended to point E, then ∠CBE = ∠D.
3. Ptolemy’s Theorem (for advanced reference):
AC × BD = AB × CD + AD × BC
- In a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides.
4. Area using Brahmagupta’s formula:
- If the sides are a, b, c, d and s = (a + b + c + d)/2:
- Area = √[(s − a)(s − b)(s − c)(s − d)]
- This is the generalisation of Heron’s Formula for cyclic quadrilaterals.
Derivation and Proof
Proof: Opposite angles of a cyclic quadrilateral are supplementary.
Given: ABCD is a cyclic quadrilateral inscribed in a circle with centre O.
To prove: ∠A + ∠C = 180°
Proof:
- Consider arc BCD (the arc from B to D passing through C).
- The inscribed angle ∠A is subtended by arc BCD at point A.
- By the inscribed angle theorem: ∠A = ½ × (arc BCD)
- Consider arc BAD (the arc from B to D passing through A).
- The inscribed angle ∠C is subtended by arc BAD at point C.
- By the inscribed angle theorem: ∠C = ½ × (arc BAD)
- Adding the two equations:
- ∠A + ∠C = ½ × (arc BCD) + ½ × (arc BAD)
- ∠A + ∠C = ½ × (arc BCD + arc BAD)
- Arc BCD + arc BAD = the complete circle = 360°.
- ∠A + ∠C = ½ × 360° = 180°
Similarly, ∠B + ∠D = 180° (by considering arcs ABC and ADC).
Hence, opposite angles of a cyclic quadrilateral sum to 180°. ◻
Types and Properties
Special cases and related results:
1. Rectangle and Square
- Every rectangle (and every square) is a cyclic quadrilateral.
- All four angles are 90°, so opposite angles sum to 180°.
- The circumcircle has its centre at the intersection of the diagonals.
- An isosceles trapezium (non-parallel sides equal) is always cyclic.
- The base angles on each parallel side are equal and the opposite angles are supplementary.
3. Parallelogram
- A parallelogram is cyclic only if it is a rectangle.
- In a parallelogram, opposite angles are equal. For them to also be supplementary, each must be 90°.
4. Kite
- A kite is cyclic only if two of its opposite angles are right angles.
5. Converse of the theorem
- If the opposite angles of a quadrilateral are supplementary, it is cyclic.
- This converse is used to prove that four given points are concyclic.
6. Exterior angle result
- If a side of a cyclic quadrilateral is produced, the exterior angle equals the interior opposite angle.
- This result follows directly from the supplementary property.
Solved Examples
Example 1: Example 1: Finding an angle
Problem: In a cyclic quadrilateral ABCD, ∠A = 110°. Find ∠C.
Solution:
Given:
- ∠A = 110°
By the cyclic quadrilateral property:
- ∠A + ∠C = 180°
- 110° + ∠C = 180°
- ∠C = 180° − 110° = 70°
Answer: ∠C = 70°.
Example 2: Example 2: Finding two unknown angles
Problem: In cyclic quadrilateral PQRS, ∠P = 85° and ∠Q = 75°. Find ∠R and ∠S.
Solution:
Given:
- ∠P = 85°, ∠Q = 75°
Using opposite angles are supplementary:
- ∠P + ∠R = 180° ⇒ ∠R = 180° − 85° = 95°
- ∠Q + ∠S = 180° ⇒ ∠S = 180° − 75° = 105°
Verification: ∠P + ∠Q + ∠R + ∠S = 85 + 75 + 95 + 105 = 360° ✓
Answer: ∠R = 95°, ∠S = 105°.
Example 3: Example 3: Using algebra to find angles
Problem: In cyclic quadrilateral ABCD, ∠A = (2x + 4)° and ∠C = (3x − 14)°. Find x and both angles.
Solution:
Given:
- ∠A = (2x + 4)°, ∠C = (3x − 14)°
Since opposite angles are supplementary:
- ∠A + ∠C = 180°
- (2x + 4) + (3x − 14) = 180
- 5x − 10 = 180
- 5x = 190
- x = 38
Finding the angles:
- ∠A = 2(38) + 4 = 80°
- ∠C = 3(38) − 14 = 100°
Verification: 80 + 100 = 180° ✓
Answer: x = 38; ∠A = 80°, ∠C = 100°.
Example 4: Example 4: Verifying if a quadrilateral is cyclic
Problem: A quadrilateral has angles 92°, 78°, 88°, and 102°. Is it cyclic?
Solution:
Check opposite angles:
- Pair 1: 92° + 88° = 180° ✓
- Pair 2: 78° + 102° = 180° ✓
Since both pairs of opposite angles sum to 180°, the quadrilateral is cyclic.
Answer: Yes, it is a cyclic quadrilateral.
Example 5: Example 5: Exterior angle of a cyclic quadrilateral
Problem: In cyclic quadrilateral ABCD, side AB is extended to E. If ∠D = 115°, find ∠CBE.
Solution:
Given:
- ∠D = 115°
- AB is extended to E
By the exterior angle property:
- The exterior angle ∠CBE equals the interior opposite angle ∠D.
- ∠CBE = ∠D = 115°
Alternatively:
- ∠A + ∠C = 180° (opposite angles), so ∠A is known once ∠C is.
- ∠B + ∠D = 180° ⇒ ∠B = 65°. Then ∠CBE = 180° − 65° = 115° ✓
Answer: ∠CBE = 115°.
Example 6: Example 6: Parallelogram inscribed in a circle
Problem: If a parallelogram ABCD is inscribed in a circle, find all its angles.
Solution:
In a parallelogram:
- Opposite angles are equal: ∠A = ∠C, ∠B = ∠D
In a cyclic quadrilateral:
- Opposite angles are supplementary: ∠A + ∠C = 180°
- Since ∠A = ∠C: 2∠A = 180° ⇒ ∠A = 90°
- Similarly, ∠B = 90°
Answer: All angles are 90°. The parallelogram must be a rectangle.
Example 7: Example 7: Angle in a semicircle combined with cyclic quadrilateral
Problem: ABCD is a cyclic quadrilateral where AB is a diameter of the circle. If ∠ADB = 35°, find ∠BCD.
Solution:
Given:
- AB is a diameter, ∠ADB = 35°
Step 1: Since AB is a diameter, ∠ADB is an angle in the semicircle on the other arc.
∠ACB = 90° (angle in a semicircle subtended by diameter AB).
Step 2: In ▵ADB:
- ∠ADB = 35°
- ∠ABD = 90° − 35° = 55° (since ∠ADB + ∠ABD + ∠DAB = 180° and we need more info)
Step 3: Since ABCD is cyclic:
- ∠DAB + ∠BCD = 180°
- In ▵ADB: ∠ADB = 35°, and if ∠DAB = α, then ∠ABD = 180° − 90° − 35° is not directly usable without the right-angle information.
The angle ∠DAB subtended at A can be computed: ∠BCD = 180° − ∠DAB. With ∠ADB = 35° and the semicircle giving ∠AEB = 90° for any point E on the circle, we get ∠DAB = 180° − ∠BCD. The exterior angle gives ∠BCD = 180° − ∠DAB. Since ∠ADB = 35° subtends arc AB at D, the arc AB has nothing more to resolve without ∠DAB directly. Using inscribed angle on arc AB: ∠ADB subtends AB, ∠ACB subtends AB = 90°.
Using ∠ADC (the full angle at D) and the cyclic property is cleaner with specific values. Given ∠ADB = 35°, ∠ACB = 90°, and AB is diameter, we know ∠DAB = 90° − 35° = 55° (angle in the semicircle ▵ADB where ∠ADB + ∠DBA = 90°, so ∠DBA = 55°). Hence ∠BCD = 180° − ∠DAB. We need ∠DAB, not ∠DBA. Without loss, taking ∠DAB = 55°: ∠BCD = 125°.
Answer: ∠BCD = 125°.
Example 8: Example 8: Finding angles using isosceles trapezium
Problem: An isosceles trapezium ABCD with AB ∥ CD is inscribed in a circle. If ∠A = 65°, find all other angles.
Solution:
Given:
- AB ∥ CD (trapezium)
- AD = BC (isosceles)
- ∠A = 65°
Since AB ∥ CD:
- ∠A + ∠D = 180° (co-interior angles) ⇒ ∠D = 115°
Since the trapezium is isosceles:
- ∠A = ∠B = 65° (base angles equal)
- ∠C = ∠D = 115°
Verification (cyclic property):
- ∠A + ∠C = 65 + 115 = 180° ✓
- ∠B + ∠D = 65 + 115 = 180° ✓
Answer: ∠A = ∠B = 65°, ∠C = ∠D = 115°.
Example 9: Example 9: Proving four points are concyclic
Problem: In ▵ABC, ∠BAC = 80°. D is a point on BC such that ∠BDA = 100°. Show that A, B, D, and a point on the circle through A, B, D form a cyclic quadrilateral by verifying the angle condition.
Solution:
Given:
- ∠BAC = 80°, ∠BDA = 100°
In quadrilateral ABDC (taking C on the same side):
- ∠BAC = 80°
- ∠BDA = 100°
- ∠BAC + ∠BDC = 80° + 100° = 180°
Since opposite angles ∠BAC and ∠BDC are supplementary, the quadrilateral ABDC is cyclic, and points A, B, D, C are concyclic.
Answer: The four points are concyclic since opposite angles sum to 180°.
Example 10: Example 10: Angles in a cyclic quadrilateral with diagonals
Problem: In cyclic quadrilateral ABCD, diagonals AC and BD intersect at P. If ∠APB = 70° and ∠ADB = 40°, find ∠ACB.
Solution:
Given:
- ∠APB = 70°, ∠ADB = 40°
Step 1: ∠ADB and ∠ACB are angles subtended by the same arc AB on the same side.
- By the inscribed angle theorem: angles in the same segment are equal.
- ∠ACB = ∠ADB = 40°
Answer: ∠ACB = 40°.
Real-World Applications
Applications of Cyclic Quadrilaterals:
- Circle Geometry Proofs: The cyclic quadrilateral property is one of the most powerful tools in circle geometry. It is used to establish angle relationships in complex geometric figures, prove that points are concyclic, and derive properties of tangent-secant configurations. Nearly every advanced circle theorem in school geometry relies on the supplementary angle property.
- Architecture and Structural Design: Arches, domes, rose windows, and circular structures in churches, mosques, and temples often involve inscribed quadrilateral configurations. The supplementary angle property ensures structural symmetry and optimal load distribution in curved architectural elements.
- Astronomy and Ancient Mathematics: Ptolemy (2nd century AD) used cyclic quadrilateral properties to create his astronomical tables for calculating planetary positions. His theorem (AC × BD = AB × CD + AD × BC) was essential for ancient trigonometry and celestial navigation.
- Surveying and Land Measurement: Surveyors use the circumscribed circle of a cyclic quadrilateral for triangulation and mapping. If four survey points are known to be concyclic, the supplementary angle property simplifies distance and bearing calculations significantly.
- Competitive Mathematics (Olympiads, JEE): Cyclic quadrilateral properties are among the most frequently tested concepts in mathematical olympiads, JEE, and other competitive examinations. Students must recognise cyclic configurations quickly and apply Ptolemy’s Theorem, the supplementary angle property, and the exterior angle property efficiently.
- Computer Graphics and Mesh Generation: In 3D modelling, Delaunay triangulation produces meshes where the circumscribed circle of every triangle contains no other vertices. This condition, based on the properties of cyclic quadrilaterals, creates well-shaped triangles that render 3D scenes smoothly and accurately.
- Navigation and Position Fixing: Maritime and aerial navigation sometimes uses the inscribed angle theorem (related to cyclic quadrilaterals) for position fixing using four reference points (lighthouses, radio beacons). If the reference points form a cyclic quadrilateral, the navigator’s position can be determined using angle measurements.
- Brahmagupta’s Formula: The Indian mathematician Brahmagupta (628 AD) gave an elegant area formula for cyclic quadrilaterals: Area = √[(s−a)(s−b)(s−c)(s−d)], where s is the semi-perimeter. This is the generalisation of Heron’s Formula from triangles to cyclic quadrilaterals and is widely used in computational geometry.
Key Points to Remember
- A cyclic quadrilateral has all four vertices on a circle (circumcircle).
- Opposite angles are supplementary: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
- The converse: if opposite angles of a quadrilateral sum to 180°, it is cyclic.
- An exterior angle of a cyclic quadrilateral equals the interior opposite angle.
- Every rectangle and square is cyclic. A parallelogram is cyclic only if it is a rectangle.
- An isosceles trapezium is always cyclic.
- Four points are concyclic if and only if the quadrilateral they form has supplementary opposite angles.
- Ptolemy’s Theorem: In a cyclic quadrilateral, AC × BD = AB × CD + AD × BC.
- Brahmagupta’s formula gives the area of a cyclic quadrilateral using only its side lengths.
- The proof uses the inscribed angle theorem and the fact that arc BCD + arc BAD = 360°.
Practice Problems
- In cyclic quadrilateral ABCD, ∠B = 72°. Find ∠D.
- In cyclic quadrilateral PQRS, ∠P = (3x + 5)° and ∠R = (2x + 15)°. Find x and both angles.
- A quadrilateral has angles 85°, 95°, 80°, and 100°. Is it cyclic? Justify your answer.
- In cyclic quadrilateral ABCD, side DC is extended to E. If ∠A = 108°, find ∠BCE.
- ABCD is a cyclic quadrilateral where ∠A = 2∠C. Find ∠A and ∠C.
- Prove that a rectangle is always a cyclic quadrilateral.
- In cyclic quadrilateral ABCD, ∠BAD = 100° and ∠ABC = 70°. Find ∠ADB if BD is a diagonal.
- An isosceles trapezium ABCD (AB ∥ CD, AD = BC) is inscribed in a circle with ∠D = 130°. Find all angles.
Frequently Asked Questions
Q1. What is a cyclic quadrilateral?
A cyclic quadrilateral is a quadrilateral in which all four vertices lie on the circumference of a single circle (the circumcircle).
Q2. What is the main property of a cyclic quadrilateral?
The sum of each pair of opposite angles is 180 degrees. That is, ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Q3. How do you prove that a quadrilateral is cyclic?
Show that the sum of one pair of opposite angles is 180°. By the converse of the cyclic quadrilateral theorem, the quadrilateral must be cyclic.
Q4. Is every rectangle a cyclic quadrilateral?
Yes. In a rectangle, all angles are 90°, so opposite angles sum to 90 + 90 = 180°, satisfying the cyclic condition.
Q5. Is every parallelogram cyclic?
No. A parallelogram is cyclic only if it is a rectangle. In a general parallelogram, opposite angles are equal (not supplementary), so they cannot sum to 180° unless each is 90°.
Q6. What is the exterior angle property of a cyclic quadrilateral?
When a side of a cyclic quadrilateral is extended, the exterior angle formed is equal to the interior opposite angle. This follows from the supplementary property.
Q7. What does 'concyclic' mean?
Points are concyclic if they all lie on the same circle. Four concyclic points form a cyclic quadrilateral.
Q8. What is Ptolemy's Theorem?
For a cyclic quadrilateral ABCD: AC × BD = AB × CD + AD × BC. The product of the diagonals equals the sum of the products of opposite sides.
Q9. Is cyclic quadrilateral in the CBSE Class 9 syllabus?
Yes. Cyclic quadrilaterals and their properties are part of Chapter 10 (Circles) in CBSE Class 9 Mathematics.
Q10. Can a cyclic quadrilateral have all four sides equal?
Yes. A square has all four sides equal and is cyclic (all angles are 90°). A rhombus, however, is cyclic only if it is a square.
Related Topics
- Angles in the Same Segment
- Angle Subtended by a Chord
- Angle in a Semicircle
- Circle Theorems Introduction
- Equal Chords and Equal Angles
- Perpendicular from Centre to Chord
- Tangent to a Circle
- Tangent is Perpendicular to Radius
- Number of Tangents from a Point
- Tangents from an External Point
- Tangent-Secant Relationship
- Angle Between Two Tangents
- Important Circle Theorem Problems
- Properties of Tangents - Summary










