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

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

Proof:

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.