PQ is a diameter of a circle with centre O. The chords PS and QR produced meet externally at A. Prove that SOR PAQ = 180°.
Answer: Let ∠APQ = x and ∠AQP = y. Then ∠QAR = 180 - y and ∠PSA = 180 - x. Also, note that ∠QOS = 2y and ∠SOR = 90 - y. Then, using the exterior angle theorem in triangle APS, we have ∠SAR = x + y, and ∠ROQ = 180 - (x + y). Finally, we have: ∠SOR + ∠ROQ + ∠QOS = (90 - y) + (180 - (x + y)) + 2y = 270 - x But we know that ∠SOR + ∠ROQ + ∠QOS = 360 (angle sum property). Therefore, we have: 270 - x = 360 Solving for x, we get x = 90. Therefore, ∠APQ = 90 and ∠AQP = 90, which implies that PAQ is a straight line, and hence SOR PAQ = 180°.