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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.