Let AB is diameter, PQ and RS are tangents drawn at ends of diameter.
We Know that the radius is perpendicular to tangent. therefore, OP ⊥PQ.
∠OAR = 90° and OAS = 90°
∠OBP = 90° and ∠OBQ = 90°
From the above, we have
∠OAR = ∠OBQ [Alternate angles]
∠OAS = ∠OBP [Alternate angles]
Since, alternate angles are equal. Hence, PQ is parallel to PS.