Question

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

Calculate:

1. ∠DAB,
2. ∠DBA,
3. ∠DBC,
4. ∠ADC.

Also, show that the ΔAOD is an equilateral triangle.

Answer 1

i. ABCD is a cyclic quadrilateral

∴ ∠DCB + ∠DAB = 180°

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

`=>` ∠DAB = 180° – 120° = 60°

ii. ∠ADB = 90°

(Angle in a semicircle is a right angle)

∴ ∠DBA = 90° – ∠DAB

= 90° – 60°

= 30°

iii. OD = OB

∴ ∠ODB = ∠OBD

Or ∠ABD = 30°

Also, AB || ED

∴ ∠DBC = ∠ODB = 30° (Alternate angles)

iv. ∠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.