AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter, let us prove that OA ∥ RQ

In Δ APO and Δ AQO

AP = AQ (Tangents drawn to the same circle from an external point are always equal in length)

AO = AO (Common)

∠ OPA = ∠ OQA (Tangents are perpendicular to the line joining the centre)

So Δ APO and Δ AQO are congruent by S.A.S. axiom of congruency.

So from corresponding parts of congruent triangle we get

∠ POA = ∠ QOA = x (Let)

∠ POA + ∠ QOA = 2x

∠ QOR = 180⁰-2x (Angle on a straight line)

∠ OQR = ∠ ORQ (Δ OQR is isosceles)

∠ OQR + ∠ ORQ = 180⁰-(180⁰-2x) = 2x

Since they are equal so

∠ OQR = x = ∠ AOQ

∠ OQR and ∠ AOQ forms a pair of alternate interior angles

So OA∥ RQ