Exercise 12.5 — Cyclic Quadrilaterals
Cyclic quadrilaterals and their properties.
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.
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.
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.
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°).
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°:
Quick Answer Summary — Question 1
| Part | Key Theorem Used | x° | y° |
|---|---|---|---|
| (i) | Triangle angle sum (180°) + symmetry | 75° | 75° |
| (ii) | Opposite angles in cyclic quadrilateral sum to 180° | 70° | 95° |
| (iii) | Angle in semicircle = 90° + triangle sum | 90° | 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 + ∠C = 180° AND ∠B + ∠D = 180° ⟺ ABCD is a Cyclic QuadrilateralQuestion 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 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.
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).
| Figure | Shape | Can Be Inscribed? | Reason |
|---|---|---|---|
| (a) | Rectangle | ✔ YES | Opposite angles: 90° + 90° = 180° ✓ |
| (b) | Isosceles Trapezium | ✔ YES | Isosceles trapezium always has supplementary opposite angles |
| (c) | Obtuse Triangle | ✔ YES | Every triangle has a circumscribed circle |
| (d) | Non-rectangular Parallelogram | ✗ NO | Opposite angles are equal, not supplementary — cannot be cyclic |
| (e) | Acute Isosceles Triangle | ✔ YES | Every triangle can be inscribed in its circumcircle |
| (f) | Quadrilateral PQRS (PR as diameter) | ✔ YES | ∠PQR = ∠PSR = 90° by angle in semicircle theorem |
Key Theorems Used in This Exercise
| Theorem | Statement | Used 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.
Quick Reference — All Answers at a Glance
| Question | Part | Answer / Conclusion |
|---|---|---|
| Q1 | (i) | x° = 75°, y° = 75° |
| Q1 | (ii) | x° = 70°, y° = 95° |
| Q1 | (iii) | x° = 90°, y° = 40° |
| Q2 | — | D lies on the circle (ABCD is cyclic since ∠B + ∠D = 180°) |
| Q3 | — | A cyclic parallelogram is a Rectangle (all angles = 90°) |
| Q4 | — | A 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.