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Question
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. show that BC = BD.
Solution
Join OC,
∠BCD = ∠BAC = 30° ...(Angles in alternate segment)
Arc BC subtends ∠DOC at the centre of the circle and ∠BAC at the remaining part of the circle.
∴ ∠BOC = 2∠BAC = 2 × 30° = 60°
Now in ΔOCD,
∠BOC or ∠DOC = 60°
∠OCD = 90° ...(OC ⊥ CD)
∴ ∠DCO + ∠ODC = 90°
`=>` 60° + ∠ODC = 90°
`=>` ∠ODC = 90° – 60° = 30°
Now in ΔBCD,
∵ ∠ODC or ∠BDC = ∠BCD = 30°
∴ BC = BD
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