Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
Given: Quadrilateral ABCD circumscribes a circle with centre O, touching sides AB, BC, CD, DA at P, Q, R, S respectively.
To prove: ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°
Proof: Join OP, OQ, OR, OS.
Since tangents from an external point are equal:
- From A: AP = AS → ∠AOP = ∠AOS
- From B: BP = BQ → ∠BOP = ∠BOQ
- From C: CQ = CR → ∠COQ = ∠COR
- From D: DS = DR → ∠DOS = ∠DOR
Since ∠AOP + ∠AOS + ∠BOP + ∠BOQ + ∠COQ + ∠COR + ∠DOS + ∠DOR = 360°
We get: 2(∠AOP + ∠BOP + ∠COR + ∠DOR) = 360°
∴ (∠AOB) + (∠COD) = 180°
Similarly, ∠AOD + ∠BOC = 180° [Hence proved]
Source: Chapter 10, Section 10.3 (Exercise 10.2, Q.13)
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Explanation
- The key idea is that tangents from an external point are equal, so each vertex creates two equal angles at O (by congruent triangles using RHS or the property from Theorem 10.2).
- All eight angles around O sum to 360°. Grouping them by opposite sides gives two pairs each summing to 180°.
- Examiners expect: clear labelling of points of contact, invoking the equal-tangent property for each vertex, and the angle-sum step. Don't skip the grouping step.