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Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

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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.

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