Given: CD and EF are two parallel tangents at points A and B of a circle with centre O.
To prove: AB passes through centre O or AOB is a diameter of the circle.
Construction: Join OA and OB. Draw OM || CD
Since, OM || CD,
OM || AC
We know that sum of adjacent interior angles is 180°.
⇒ ∠CAO + ∠MOA = ∠1 + ∠2 = 180°
We know that a tangent to a circle is perpendicular to the radius through the point of contact.
∠CAO = 90°
⇒ 90° + ∠MOA = 180°
⇒ ∠MOA = 90°
Similarly, ∠3 = ∠MOB = 90°
∴ ∠MOA + ∠MOB = 90° + 90° = 180°
Thus, AOB is a straight line passing through O.
Ans. Hence, the line segment joining the points of contact of two parallel tangents of a circle passes through its centre.
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