In cyclic quadrilateral ABCD; AD = BC, ∠BAC = 30° and ∠CBD = 70°; find :

(i) ∠BCD

(ii) ∠BCA

(iii) ∠ABC

(iv) ∠ADC

(i) From figure,

∠DAC = ∠CBD = 70° [Angles in same segment are equal]

∠BAD = ∠BAC + ∠DAC = 30° + 70° = 100°.

Since, sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠BAD + ∠BCD = 180°

⇒ 100° + ∠BCD = 180°

⇒ ∠BCD = 180° - 100° = 80°.

Hence, ∠BCD = 80°.

(ii) Since, AD = BC

∠ACD = ∠BDC …………(1) [Equal chords subtends equal angles]

As,

Angles in same segment are equal.

∠ACB = ∠ADB ………..(2)

Adding (1) and (2) we get,

⇒ ∠ACD + ∠ACB = ∠BDC + ∠ADB

⇒ ∠BCD = ∠ADC = 80°.

In △BCD,

⇒ ∠CBD + ∠BCD + ∠BDC = 180°

⇒ 70° + 80° + ∠BDC = 180°

⇒ 150° + ∠BDC = 180°

⇒ ∠BDC = 180° - 150° = 30°.

∠ACD = ∠BDC = 30° [As equal chords subtends equal angles.]

From figure,

⇒ ∠BCA = ∠BCD - ∠ACD = 80° - 30° = 50°.

Hence, ∠BCA = 50°.

(iii) As sum of opposite angles in a cyclic quadrilateral = 180°.

⇒ ∠ADC + ∠ABC = 180°

⇒ 80° + ∠ABC = 180°

⇒ ∠ABC = 180° - 80° = 100°.

Hence, ∠ABC = 100°.

(iv) From part (ii) we get,

Hence, ∠ADC = 80°.