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Question
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠ADC
Also, show that the ΔAOD is an equilateral triangle.
Solution
∠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.
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