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* = 2*r* = Diameter

Hence AB is the diameter of the circle.

**
**