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

(i) ∠DAB,

(ii) ∠DBA,

(iii) ∠DBC,

(iv) ∠ADC.

Also, show that the △AOD is an equilateral triangle.

(i) ABCD is a cyclic quadrilateral.

∴ ∠DCB + ∠DAB = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ ∠DAB = 180° - ∠DCB

⇒ ∠DAB = 180° - 120° = 60°.

Hence, ∠DAB = 60°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∠ADB = 90°.

In △DAB,

⇒ ∠ADB + ∠DAB + ∠DBA = 180°

⇒ 90° + 60° + ∠DBA = 180°

⇒ 150° + ∠DBA = 180°

⇒ ∠DBA = 180° - 150° = 30°.

Hence, ∠DBA = 30°.

(iii) OD = OB (Radius of circle)

∴ ∠ODB = ∠OBD [Angles opposite to equal sides are equal]

From figure,

∠OBD = ∠DBA = 30°.

∴ ∠ODB = 30°.

As, DO || BC

∴ ∠DBC = ∠ODB = 30° [Alternate angles are equal.]

Hence, ∠DBC = 30°.

(iv) From figure,

∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°.

In cyclic quadrilateral ABCD,

∴ ∠ADC + ∠ABC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ ∠ADC = 180° - ∠ABC

⇒ ∠ADC = 180° - 60° = 120°.

Hence, ∠ADC = 120°.

In △AOD,

OA = OD [Radius of same circle]

∠AOD = ∠DAO [Angles opposite to equal sides are equal]

From figure,

⇒ ∠DAO = ∠DAB = 60°.

∴ ∠AOD = ∠DAO = ∠ADO = 60°

Hence, proved that △AOD is an equilateral triangle.