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.