Detailed Answer :
Let AB be the diameter of a circle, with centre O. The tangents PQ and RS are drawn at points A and B, respectively.
We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA ⊥ RS and OB ⊥ PQ
⇒ ∠ OAR = 90°
∠ OAS = 90°
∠ OBP = 90°
∠ OBQ = 90°
We can observe the following :
∠ OAR = ∠ OBQ and ∠ OAS = ∠ OBP
Also, these are the pair of alternate interior angles. Since, alternate interior angles are equal, the lines PQ and RS are parallel to each other.
