#### Question

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

Calculate:

(i) ∠DAB, (ii) ∠DBA, (iii) ∠DBC, (iv) ∠ADC.

Also show that the ΔAOD is an equilateral triangle .

#### Solution

(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)]

∠DAO = 60°

⇒ ∠ADO = ∠AOD =∠DAO = 60°

∴ ∆AOD is an equilateral triangle.

Is there an error in this question or solution?

Solution In the Given Figure, Ab is a Diameter of the Circle with Centre O. Do is Parallel to Cb and ∠Dcb = 120°. Calculate: (I) ∠Dab, (Ii) ∠Dba, (Iii) ∠Dbc, (Iv) ∠Adc. Also Show that the δAod is an Equilatera Concept: Arc and Chord Properties - Angle in a Semi-circle is a Right Angle.