In the given figure, equilateral ∆ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then, ∠BDC = ?
(a) 90°
(b) 60°
(c) 120°
(d) 150°

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Solution

(c) 120°
Since ΔABC is an equilateral triangle, each of its angle is 60°.
∴ ∠BAC = 60°
In a cyclic quadrilateral ABCD, we have:
∠BAC + ∠BDC = 180°
⇒ 60° + ∠BDC = 180°
⇒ ∠BDC = (180° - 60°) = 120°

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