ABCD is a cyclic quadrilateral, whose diagonals intersect at a point E. If ∠DBC = 80° and ∠BAC = 40°, then find ∠BCD. Further, if AB = BC, then find ∠ECD.

Given, in quadrilateral, ∠BAC = 40°

and ∠DBC = 80°

here, ∠CDB= ∠BAC= 40°

[∴ angles in the same segment of a circle are equal]

In ∆BCD, ∠DBC + ∠BCD + ∠CDB = 180°

[ ∴ sum of all angles of a triangle is 180° ]

80° + ∠BCD + 40° = 180° [from Eq. (i)]

∠BCD = 180° -40° - 80°

∠BCD = 60° …(ii)

In AABC, AB = BC [given][ ∴

[ ∴angles opposite to equal sides of a triangle are equal]

∠BCA = 40° [from Eq. (i)] …(iii)

From Eq. (ii), we have ∠ BCD = 60°

∠BCA + ∠ACD = 60°

40° + ∠ACD = 60° [from Eq. (iii)]

∴ ∠ACD = 20° or ∠ECD = 20°