Question

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°. 

Calculate: ∠ADC 

Also, show that the ΔAOD is an equilateral triangle.

Answer 1

∠ABD + ∠DBC = 30° + 30° = 60° 

⇒ ∠ABC = 60°

In cyclic quadrilateral ABCD,

∠ADC + ∠ABC = 180°

(Pair of opposite angles in a cyclic quadrilateral are supplementary)

⇒ ∠ADC = 180° – 60° = 120°

In ∆AOD, OA = OD        ...(Radii of the same circle)

∠AOD = ∠DAO Or ∠DAB = 60°    ...[Proved in (i)]

∠AOD = 60°

⇒ ∠ADO = ∠AOD = ∠DAO = 60°

∴ ∆AOD is an equilateral triangle.