Let AB is diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively. Radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = 90°
∠OAS = 90°
∠OBP = 90°
∠OBQ = 90°
It can be observed that
∠OAR = ∠ OBQ (Alternate interior angles)
∠OAS = ∠ OBP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
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