Question

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

If $\angle BCD={120}^{\circ },$ calculate

(i) $\angle$BAD, (ii) $\angle$ABD,

(iii) $\angle$CBD, (iv) $\angle$ADC.

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

Answer 1

ANSWER:
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 °
∴ ∠CB D = 30 °
(iv)
∠ ADC = ∠ ADB + ∠ CDB
= 90 ° + 30 ° = 120 °
In Δ AOD, we have:
∠ ODA + ∠ OAD + ∠ AOD = 180 °
⇒ 60 ° + 60 ° + ∠ AOD = 180 °
⇒ ∠ AOD = 180 ° – 120 ° = 60 °
Since all the angles of Δ AOD are of 60 ° each, Δ AOD is an equilateral triangle.