∠CAD = ∠DBC= 70° [Angles in the same segment]

Therefore, ∠DAB = ∠CAD + ∠BAC

= 70° + 30° = 100°

But, ∠DAB + ∠BCD = 180°

[Opposite angles of a cyclic quadrilateral]

So, ∠BCD = 180° – 100° = 80°

Now, we have AB = BC

Therefore, ∠BCA = 30° [Opposite angles of an isosceles triangle]

Again, ∠DAB + ∠BCD = 180°

[Opposite angles of a cyclic quadrilateral]

⇒ 100° + ∠BCA + ∠ECD = 180° [∵∠BCD = ∠BCA + ∠ECD]

⇒ 100° + 30° + ∠ECD = 180°

⇒ 130° + ∠ECD = 180°

⇒ ∠ECD = 180° – 130° = 50°

Hence, ∠BCD = 80° and ∠ECD = 50° **Ans.**