In the given figure, O is the center of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find :

(i) ∠BOD

(ii) ∠BPD

(i) From figure,

∠BCD = ∠ABC = 55° [Alternate angles are equal.]

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOD = 2∠BCD = 2 x 55° = 110°.

Hence, ∠BOD = 110°.

(ii) We know that,

A tangent line is always at a right angle to the radius of the circle at the point of tangency.

∴ ∠OBP = 90° and ∠ODP = 90°.

In quadrilateral ODPB,

⇒ ∠BOD + ∠OBP + ∠ODP + ∠BPD = 360° [Angle sum property of quadrilateral]

⇒ 110° + 90° + 90° + ∠BPD = 360°

⇒ ∠BPD = 360° - 290° = 70°.

Hence, ∠BPD = 70°.