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

Given: A circle with center O and AB and CD are two parallel tangents at points P and Q on the circle.

To Prove: PQ passes through O

Construction: Draw a line EF parallel to AB and CD and passing through O

Proof :

∠OPB = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

Now, AB || EF

∠OPB + ∠POF = 180°

90° + ∠POF = 180°

∠POF = 90° …[1]

Also,

∠OQD = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

Now, CD || EF

∠OQD + ∠QOF = 180°

90° + ∠QOF = 180°

∠QOF = 90° [2]

Now From [1] and [2]

∠POF + ∠QOF = 90 + 90 = 180°

So, By converse of linear pair POQ is a straight Line

i.e. O lies on PQ

Hence Proved.

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