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ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC = 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

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∠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.

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