Question

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.

Answer 1

∠CDB = ∠BAC = $30\circ$ .....(i) (Angles in the same segment of a circle are equal)

∠DBC = 70∘ .....(ii)

In ΔBCD,

∠BCD + ∠DBC + ∠CDB = 180∘ (Sum of all angles of a triangle is 180∘)

∠BCD + 70∘ + 30∘ = 180∘ (using (i) and (ii))

∠BCD = 180∘ - 100∘ = 80∘ .....(iii)

In ΔABC,

Given: AB= BC

So, ∠BCA = ∠BAC = 30∘.....(iv) (Angles opposite to equal sides of a triangle are equal)

Now, ∠BCD = 80∘ from (iii)

∠BCA + ∠ECD = 80∘

30∘ + ∠ECD = 80∘

∴ ∠ECD = 50∘