∠CDB = ∠BAC = 300 ...(i) [Angles in the same segment of a circle are equal]
∠DBC = 700 ....(ii)
In ∆BCD,
∠BCD + ∠DBC + ∠CDB = 1800
[Sum of all he angles of a triangle is 1800 ]
⇒ ∠BCD + 700 + 30.0 = 1800 [Using (i) and (ii)]
⇒ ∠BCD + 1000 = 1800
⇒ ∠BCD = 1800 - 1000
⇒ ∠BCD = 800 ...(iii)
In ∆ABC, AB = BC
∴ ∠BCA = ∠BAC = 300 ...(iv)
[Angles opposite to equal sides of a triangle are equal]
Now, ∠BCD = 800 [From (iii)]
⇒ ∠BCA + ∠ECD = 800
⇒ 300 + ∠ECD = 800
⇒ ∠ECD = 800 - 300
⇒ ∠ECD = 500