Exercise 12.5 — Cyclic Quadrilaterals

Cyclic quadrilaterals and their properties.

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Exercise 12.5 — Cyclic Quadrilaterals and Circles

Exercise 12.5 is part of Chapter 12, Circles, from Class 9 Mathematics for students following the CBSE, Telangana, and Andhra Pradesh board syllabi. This exercise focuses on one of the most important concepts in circle geometry: cyclic quadrilaterals — quadrilaterals whose four vertices all lie on a single circle. The problems in this exercise develop both your proof-writing skills and your understanding of angle relationships in circles.

A quadrilateral is called cyclic (or concyclic) when all four of its vertices touch the circumference of the same circle. The key property that defines a cyclic quadrilateral is simple but powerful: opposite angles always add up to 180°. This single fact is the engine behind every question in this exercise.

Cyclic Quadrilateral Angle in Semicircle Cyclic Rectangle Cyclic Rhombus = Square Inscribed Polygons
KEY THEOREM
In a cyclic quadrilateral, opposite angles are supplementary (add up to 180°). That is, ∠A + ∠C = 180° and ∠B + ∠D = 180°.
💡 Why it matters: This theorem is tested repeatedly in board exams (Telangana SSC, AP SSC, CBSE Class 9). Questions 2, 3, and 4 in this exercise all rely on the supplementary opposite-angles property.

Question 1 — Find the Values of x° and y°

This question shows three figures, each involving a cyclic quadrilateral (or a related angle in a circle). The key theorems used are: opposite angles in a cyclic quadrilateral sum to 180°, and the angle in a semicircle equals 90°. Let's work through each figure carefully.

Part (i)
Figure with angles x°, y°, and 30°

The figure shows a cyclic quadrilateral where one angle is 30°, and two other angles are x° and y°. From the figure, x° and y° are opposite to each other, and it is given that x° = y° (the figure is symmetric). Also, the three angles shown together form a triangle in the circle or relate by the inscribed angle theorem.

Given: Three angles form a triangle inscribed in the circle. Sum of angles in a triangle = 180° x° + y° + 30° = 180° ⟹ x° + y° = 180° − 30° = 150° From the figure: x° = y° (symmetric angles) ⟹ 2x° = 150° ⟹ x° = 75° ⟹ y° = 75° ∴ x° = 75°, y° = 75°
75°
Inscribed angle
75°
Equal by symmetry
Part (ii)
Cyclic quadrilateral with angles 85°, 110°, x°, and y°

This figure is a cyclic quadrilateral. Opposite angles add up to 180°. The known angles are 85° (opposite to y°) and 110° (opposite to x°).

For x°: x° and 110° are opposite angles in the cyclic quadrilateral. x° + 110° = 180° (opposite angles in cyclic quadrilateral) ⟹ x° = 180° − 110° = 70° For y°: y° and 85° are opposite angles in the cyclic quadrilateral. y° + 85° = 180° ⟹ y° = 180° − 85° = 95° ∴ x° = 70°, y° = 95°
70°
180° − 110°
95°
180° − 85°
Part (iii)
Circle with centre O, angle 50°, and angles x°, y°

The figure shows a circle with centre O. Since the angle is in a semicircle (the angle subtended by the diameter), we have x° = 90°. Then, using the triangle angle sum with the remaining angle of 50°:

Step 1: x° = 90° (angle in a semicircle = 90°, since OA = OB = radius) Step 2: In the triangle formed: x° + y° + 50° = 180° 90° + y° + 50° = 180° ⟹ y° = 180° − 90° − 50° = 40° ∴ x° = 90°, y° = 40°
90°
Angle in semicircle
40°
180° − 90° − 50°

Quick Answer Summary — Question 1

PartKey Theorem Used
(i)Triangle angle sum (180°) + symmetry75°75°
(ii)Opposite angles in cyclic quadrilateral sum to 180°70°95°
(iii)Angle in semicircle = 90° + triangle sum90°40°

Question 2 — Prove That Vertex D Lies on the Circle

This is a proof question. Vertices A, B, and C of quadrilateral ABCD already lie on a circle. We are told that ∠A + ∠C = 180°. We need to prove that D also lies on the same circle — making ABCD a cyclic quadrilateral.

A B C D C All 4 vertices on the circle
Cyclic Quadrilateral ABCD
∠A + ∠C = 180° ⟹ D lies on circle
Question 2 — Proof
Prove vertex D lies on the circle when ∠A + ∠C = 180°
Given: ∠A + ∠C = 180° Step 1: In quadrilateral ABCD, sum of all angles = 360° ∠A + ∠B + ∠C + ∠D = 360° Step 2: Rearrange by grouping opposite pairs: (∠A + ∠C) + (∠B + ∠D) = 360° Step 3: Substitute ∠A + ∠C = 180°: 180° + (∠B + ∠D) = 360° ⟹ ∠B + ∠D = 180° Conclusion: Both pairs of opposite angles are supplementary: ∠A + ∠C = 180° AND ∠B + ∠D = 180° ∴ ABCD is a cyclic quadrilateral ⟹ D lies on the same circle as A, B, C. ✓
Logic used: The converse of the cyclic quadrilateral theorem states: If both pairs of opposite angles in a quadrilateral are supplementary, then the quadrilateral is cyclic. This proof applies exactly that converse.
∠A + ∠C = 180° AND ∠B + ∠D = 180° ⟺ ABCD is a Cyclic Quadrilateral

Question 3 — A Cyclic Parallelogram Is a Rectangle

This question asks us to prove that if a parallelogram is inscribed in a circle (i.e., it is cyclic), then it must be a rectangle. The proof uses two properties simultaneously: the opposite angle property of a parallelogram, and the supplementary opposite angle property of a cyclic quadrilateral.

Question 3 — Proof
If a parallelogram is cyclic, then it is a rectangle
Property 1 — Parallelogram: In parallelogram ABCD, opposite angles are equal. ∠A = ∠C …(1) Property 2 — Cyclic quadrilateral: In cyclic quadrilateral ABCD, opposite angles are supplementary. ∠A + ∠C = 180° …(2) Combining (1) and (2): From (1): ∠C = ∠A → substitute into (2): ∠A + ∠A = 180° 2∠A = 180° ∠A = 90° ⟹ ∠C = 90° Similarly: ∠B = ∠D and ∠B + ∠D = 180° ⟹ ∠B = 90° and ∠D = 90° ∴ All four angles = 90° ⟹ The parallelogram is a Rectangle. ✓
Key idea: A parallelogram has equal opposite angles, but a cyclic quadrilateral forces opposite angles to be supplementary. The only value that is both equal to its opposite AND supplementary to it is 90°. So every angle must be a right angle, making the shape a rectangle.
📌 Board exam tip (Telangana & AP SSC): This type of "if … then it is a rectangle" proof frequently appears as a 4-mark question. The two-property approach — using both the parallelogram property and the cyclic property together — is the expected method.
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Question 4 — A Cyclic Rhombus Is a Square

A rhombus is a parallelogram with all four sides equal. If such a rhombus is inscribed in a circle (cyclic), we must prove it becomes a square. The proof is almost identical to Question 3, because a rhombus is a special parallelogram.

A B C D 90° 90° 90° 90°
Cyclic Rhombus = Square
All sides equal + All angles 90°
Question 4 — Proof
Prove a cyclic rhombus is a square
A rhombus is a parallelogram. So, opposite angles of the rhombus are equal: ∠A = ∠C …(1) The rhombus is also cyclic (inscribed in a circle). So, opposite angles are supplementary: ∠A + ∠C = 180° …(2) Using (1) in (2): ∠A + ∠A = 180° ⟹ 2∠A = 180° ⟹ ∠A = 90° ∴ ∠C = 90° Similarly: ∠B = ∠D and ∠B + ∠D = 180° ⟹ ∠B = 90° and ∠D = 90° The rhombus already has all four sides equal. Now it also has all four angles = 90°. ∴ A cyclic rhombus is a Square. ✓
Why a square and not just a rectangle? A rectangle has right angles but sides need not be equal. A rhombus has equal sides but angles need not be 90°. When a rhombus becomes cyclic, the angles are forced to 90°. Now it has both equal sides AND right angles — making it a square by definition.
Rhombus (all sides equal) + Cyclic (∠A + ∠C = 180°) = Square (all sides equal AND all angles 90°)

Question 5 — Which Polygons Can Be Inscribed in a Circle?

The question asks you to draw a circle and inscribe each given polygon. If a polygon cannot be inscribed in a circle, write "Not Possible." A polygon can be inscribed in a circle only if all its vertices lie on a single circle — which requires the opposite angles to be supplementary (in the case of quadrilaterals).

P Q R S
(a) Rectangle
✔ Possible
Opposite angles = 90° + 90° = 180° ✓ All four vertices lie on circle.
A B C D Isosceles trapezium
(b) Trapezium
✔ Possible
An isosceles trapezium can be inscribed (opposite angles supplementary). A general trapezium cannot.
A B C
(c) Obtuse Triangle
✔ Possible
Any triangle can be inscribed in a circle (its circumscribed circle). The obtuse angle vertex will lie on the minor arc.
✗ NOT cyclic
(d) Non-rectangular Parallelogram
✗ Not Possible
Opposite angles of a parallelogram are equal, not supplementary (unless 90°). So it cannot be cyclic unless it is a rectangle.
Isosceles, all angles <90°
(e) Acute Isosceles Triangle
✔ Possible
Any triangle, including acute isosceles, can be inscribed in its circumscribed circle.
P Q R S diameter
(f) Quadrilateral PQRS (PR = diameter)
✔ Possible
With PR as diameter, ∠PQR = ∠PSR = 90° (angle in semicircle). PQRS is a valid cyclic quadrilateral.
FigureShapeCan Be Inscribed?Reason
(a)Rectangle✔ YESOpposite angles: 90° + 90° = 180° ✓
(b)Isosceles Trapezium✔ YESIsosceles trapezium always has supplementary opposite angles
(c)Obtuse Triangle✔ YESEvery triangle has a circumscribed circle
(d)Non-rectangular Parallelogram✗ NOOpposite angles are equal, not supplementary — cannot be cyclic
(e)Acute Isosceles Triangle✔ YESEvery triangle can be inscribed in its circumcircle
(f)Quadrilateral PQRS (PR as diameter)✔ YES∠PQR = ∠PSR = 90° by angle in semicircle theorem
💡 Rule to remember: A quadrilateral can be inscribed in a circle if and only if its opposite angles add up to 180°. Triangles always qualify (every triangle has a circumscribed circle). A general parallelogram — unless it's a rectangle — always fails this test.

Key Theorems Used in This Exercise

TheoremStatementUsed In
Cyclic Quadrilateral Theorem Opposite angles of a cyclic quadrilateral sum to 180° Q1(ii), Q2, Q3, Q4, Q5
Converse Theorem If opposite angles of a quadrilateral sum to 180°, it is cyclic Q2
Angle in a Semicircle The angle subtended by a diameter at any point on the circle is 90° Q1(iii), Q5(f)
Triangle Angle Sum Sum of angles in a triangle = 180° Q1(i), Q1(iii)
Circumscribed Circle Every triangle can be inscribed in a unique circle (circumcircle) Q5(c), Q5(e)

Common Mistakes to Avoid

  • Confusing "equal" with "supplementary": In a parallelogram, opposite angles are equal. In a cyclic quadrilateral, opposite angles are supplementary. These are completely different conditions — the proof in Q3 and Q4 works precisely because combining both conditions forces each angle to be 90°.
  • Thinking any trapezium can be inscribed: Only an isosceles trapezium is cyclic. A general (non-isosceles) trapezium has unequal legs and its opposite angles may not be supplementary.
  • Forgetting Q5(d) — parallelogram: Many students think a parallelogram (like a rhombus or rectangle) can always be inscribed. Remember — only a rectangle qualifies among parallelograms, because it's the only one with supplementary opposite angles (90° + 90° = 180°).
  • Wrong theorem in Q1(iii): The angle x° = 90° comes from the "angle in a semicircle" theorem (a diameter subtends 90° at the circle), not from the cyclic quadrilateral property. Make sure you cite the correct theorem in your answer.
  • Missing the converse in Q2: The proof in Question 2 requires the converse of the cyclic quadrilateral theorem. Simply quoting the theorem itself is not enough — you must use its converse to conclude that D lies on the circle.
Exam trap: Students sometimes write "∠A + ∠C = 180° means ABCD is cyclic" without justifying why. In board exams (Telangana, AP, CBSE), you must explicitly state: "By the converse of the cyclic quadrilateral theorem, since both pairs of opposite angles are supplementary, ABCD is cyclic."

Quick Reference — All Answers at a Glance

QuestionPartAnswer / Conclusion
Q1(i)x° = 75°, y° = 75°
Q1(ii)x° = 70°, y° = 95°
Q1(iii)x° = 90°, y° = 40°
Q2D lies on the circle (ABCD is cyclic since ∠B + ∠D = 180°)
Q3A cyclic parallelogram is a Rectangle (all angles = 90°)
Q4A cyclic rhombus is a Square (equal sides + all angles = 90°)
Q5(a)Rectangle — Possible
Q5(b)Isosceles Trapezium — Possible
Q5(c)Obtuse Triangle — Possible
Q5(d)Non-rectangular Parallelogram — Not Possible
Q5(e)Acute Isosceles Triangle — Possible
Q5(f)Quadrilateral PQRS with PR as diameter — Possible

What This Exercise Prepares You For

Exercise 12.5 brings together the most important angle relationships in circle geometry. The concept of cyclic quadrilaterals and the supplementary angle theorem are directly examined in Telangana SSC, AP SSC, and CBSE Class 9 board exams, often carrying 4 to 5 marks. Mastering the proofs in Q3 and Q4 will help you handle any "prove that" question involving circles and quadrilaterals.

For deeper revision, go back to earlier exercises in this chapter: Exercise 12.1 – Introduction to Circles and Exercise 12.2 – Chords and Arcs. The inscribed angle theorem and arc properties you studied there form the foundation for the cyclic quadrilateral results proved here. These concepts also connect forward to Chapter 7 – Triangles (circumscribed circles) and set the stage for Class 10 Circle theorems involving tangents and secants.

📐 Board Exam Strategy (Telangana & AP SSC, CBSE): Questions 2, 3, and 4 are high-probability proof questions in annual exams. For each proof, follow the 3-step pattern: state the given property, apply the cyclic quadrilateral property, and use algebra to reach the conclusion. Always end with a boxed or clearly stated final result.
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