prove that line segment joining the points of contact of two parallel tangents is the diameter of the circle

Let tangent l is parallel to tangent m and A and B are its points of intersection with circle of center O and radius r.

To Prove : AB is the diameter

Let l and m meet at some point P

Then in quadrilateral AOBP

∠AOB + ∠OAP + ∠OBP + ∠APB = 360° ......(2)

We know that tangents to a circle is perpendicular to the radius

⇒ OA ⊥ l and OB ⊥ m

⇒ ∠OAP = 90° and ∠OBP = 90° .........(3)

Since l || m 

∴ ∠APB = 0° ........(4)

From (2), (3) and (4)

∠AOP + 90° + 90° + 0° = 360°

⇒ ∠AOB = 360° – 180° = 180°

∴ AB = AO + OB = r + r = 2r = Diameter 

Hence AB is the diameter of the circle.