(i) Since, ABCD is a cyclic quadrilateral

Then,

∠ABC + ∠ADC = 180^o

∠ABC + 110^o = 180^o

∠ABC = 70^o

Since, AD ‖ BC

Then,

∠DAB + ∠ABC = 180^o (Co. interior angle)

∠DAC + 50^o + 70^o = 180^o

∠DAC = 180^o – 120^o

= 60^o

(ii) ∠BAC = ∠BDC = 40^o (Angle in the same segment)

In by angle sum property

∠DBC + ∠BCD + ∠BDC = 180^o

80^o + ∠BCD + 40^o = 180^o

∠BCD = 60^o

(iii) Given that,

Quadrilateral ABCD is a cyclic quadrilateral

Then,

∠BAD + ∠BCD = 180^o

∠BAD = 80^o

In triangle ABD, by angle sum property

∠ABD + ∠ADB + ∠BAD = 180^o

70^o + ∠ADB + 80^o = 180^o

∠ADB = 30^o