ABCD is a cyclic quadrilateral whose vertices are equidistant from the point 0 (centre of the circle). If ∠COD = 120° and ∠BAC = 30°, then the measure of ∠BCD is

Given : OA = OB = OC = OD

To find : $$\angle$$BCD = ?

Solution : $$\angle$$COD + $$\angle$$BOC = 180° [Linear Pair]

=> $$\angle$$BOC = 180° - 120° = 60°

Also, $$\angle$$OBC = $$\angle$$OCB [$$\because$$ OB = OC]

In $$\triangle$$BOC

=> $$\angle$$BOC + $$\angle$$OCB + $$\angle$$OBC = 180°

=> $$\angle$$OCB = $$\frac{120°}{2} = 60° $$--------Eqn(1)

Also, $$\angle$$OAB = $$\angle$$OCD [Alternate interior angles]

=> $$\angle$$OCD = 30° ---------------Eqn(2)

Adding eqn (1) & (2), we get :

=> $$\angle$$OCB + $$\angle$$OCD = 60° + 30°

=> $$\angle$$BCD = 90°