in a cyclic quadrilateral ABCD; AD=BC, angle BAC=30 degree and angle CBD=70 degree; find angle BCA.
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ABCD is a cyclic quadrilateral whose diagonal intersect at E.

∠CBD = ∠CAD (Angles in the same segment are equal)
∠CAD = 70°
∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)
∠BCD + 100° = 180°
∠BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle are equal)
⇒ ∠BCA = 30°
But ∠BCD = 80°.
⇒ ∠BCA + ∠ACD = 80°
30° + ∠ACD = 80°
⇒ ∠ACD = 50°
⇒ ∠ECD = 50°.
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