ABCD is a cyclic quadrilateral in which:
(i) BC||AD, ∠ADC=110° and ∠BAC=50°. Find ∠DAC.
(ii) ∠DBC=80° and ∠BAC=40° Find ∠BCD.
(iii) ∠BCD=100° and ∠ABD=70°. Find ∠ADB.
(i) Since, ABCD is a cyclic quadrilateral
Then,
∠ABC + ∠ADC = 180o
∠ABC + 110o = 180o
∠ABC = 70o
Since, AD ‖ BC
Then,
∠DAB + ∠ABC = 180o (Co. interior angle)
∠DAC + 50o + 70o = 180o
∠DAC = 180o – 120o
= 60o
(ii) ∠BAC = ∠BDC = 40o (Angle in the same segment)
In by angle sum property
∠DBC + ∠BCD + ∠BDC = 180o
80o + ∠BCD + 40o = 180o
∠BCD = 60o
(iii) Given that,
Quadrilateral ABCD is a cyclic quadrilateral
Then,
∠BAD + ∠BCD = 180o
∠BAD = 80o
In triangle ABD, by angle sum property
∠ABD + ∠ADB + ∠BAD = 180o
70o + ∠ADB + 80o = 180o
∠ADB = 30o