In the given figure, AB is a diameter of a circle with centre O and DO || CB.

Question:

In the given figure, AB is a diameter of a circle with centre O and DO || CB.
If BCD = 120°, calculate
(i) ∠BAD
(ii) ∠ABD
(iii) ∠CBD
(iv) ∠ADC.
Also, show that ∆OAD is an equilateral triangle.

Solution:

We have,
AB is a diameter of the circle where O is the centre, DO || BC and BCD = 120°.
(i)
Since ABCD is a cyclic quadrilateral, we have:
BCD + BAD = 180°
⇒ 120° + BAD = 180°
⇒ BAD = (180° – 120°) = 60°
∴ BAD = 60°
(ii)
BDA = 90° (Angle in a semicircle)
In Δ ABD, we have:
BDA + BAD + ABD = 180°
⇒ 90° + 60° ABD = 180°
⇒ ABD = (180° – 150°) = 30°
∴ ABD = 30°
(iii)
OD = OA (Radii of a circle)
ODA = OAD
 = BAD 
= 60°
ODB = 90° - ODA = (90° - 60°) = 30°
Here, DO || BC (Given; alternate angles)
CBD = ODB = 30°
∴ ∠CBD = 30°
(iv)
ADC = ADB + CDB
= 90° + 30° = 120°
In ΔAOD, we have:
ODA + OAD +AOD = 180°
⇒ 60° + 60° + AOD = 180°
⇒ AOD = 180° – 120° = 60°

Since all the angles of ΔAOD are of 60° each, ΔAOD is an equilateral triangle.

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