Question

In the given figure ,ABCD is a cyclic quadrilateral, whose diagonal intersect at P such that <DBC =60° and <BAC = 40°. Find <DCB.

Answer 1

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°.