# ABCD is a cyclic

METHOD 1:
For chord CD,

CBD = CAD (Angles on the same segment are equal)

= 30o + 70o

= 100o

BCD + 100o = 180o

BCD = 80o

Now, In ΔABC,

AB = BC (Given)

∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)

BCA = 30°

We have,

BCD = 80°

BCA + ACD = 80°

30° + ACD = 80°

ACD = 50°

As ACD = ∠ECD

∠ECD = 50°
METHOD 2:
∠ABD = ∠ECD.....(1)  [Since angles on the same segment are equal]
In ΔABC,
Given: AB = BC
Therefore,
∠BAC = ∠BCA = 30º
Sum of angles of a triangle = 180º
30º + 30º + ∠ABC = 180º
∠ABC = 120º.....(2)
And we can see that
∠ABC = ∠ABD + ∠DBC
120º = ∠ABD + 70º
∠ABD = 50º
And from equation 1, we can see that,
∠ECD = 50º

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