Let O is the centre of the circle.
Let PQ and RS are two parallel tangents which touches the circle at points A and B.
OA and OB are joined.
To Prove : AB passes through point O.
Draw OC || RS ||PQ
Now,
OA = OB = Radius and
PQ is tangent of circle passing through A
OA ⊥ PQ
Which implies, ∠OAP = 90°
RS is the tangent of circle passing through B
OB ⊥ RS
Which implies, ∠OBR = 90°
Since PQ || OC
∠AOC + ∠OAP = 180° (Co-interior angles)
∠AOC + 90° = 180°
∠AOC = 180° – 90° = 90°
Similarly, ∠BOC = 90°
∠AOC + ∠BOC = 90° + 90° = 180°
AOB is a straight line. Therefore, AB passes through the centre of the circle.