prove that the tangents drawn at the end of the diameter are parallel
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Here AB is a diameter of the circle with centre O, two tangents PQ and RS drawn at points A and B respectively.
Radius will be perpendicular to these tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = ∠OAS = ∠OBP = ∠OBQ = 90º
Therefore,
∠OAR = ∠OBQ (Alternate interior angles)
∠OAS = ∠OBP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Radius will be perpendicular to these tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = ∠OAS = ∠OBP = ∠OBQ = 90º
Therefore,
∠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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