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

Consider the diagram,

Consider triangle  ABD and BCD, ∠CAD = ∠CBD = 70°. (Angles in the same segment are equal)

Hence, ∠BAD = ∠CAB + ∠DAC

= 30° + 70°  = 100°

Thus, ∠BAD = 100°

Since ABCD is a cyclic Quadrilateral, 

We know that,

The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

∠BAD + ∠BCD = 180°

∠BCD = 180° - 100°

= 80°

Thus, ∠BCD = 80°

Also given AB = BC.

So, ∠BCA = ∠BAC = 30° ( angles of isosceles triangle are equal)

∠ECD = ∠BCD - ∠BCA

= 80° - 30°

= 50°

Thus, ∠ECD = 50°