Question:
In the given figure, AOB is a diameter of a circle and CD || AB. If ∠BAD = 30°, then ∠CAD = ?
(a) 30°
(b) 60°
(c) 45°
(d) 50°
Solution:
(a) 30°
∠ADC = ∠BAD = 30° (Alternate angles)
∠ADB = 90° (Angle in semicircle)
∴ ∠CDB = (90° + 30°) = 120°
But ABCD being a cyclic quadrilateral, we have:
∠BAC + ∠CDB = 180°
⇒ ∠BAD + ∠CAD + ∠CDB = 180°
⇒ 30° + ∠CAD + 120° = 180°
⇒ ∠CAD = (180° - 150°) = 30°