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Question
In Figure, AB is diameter and AC is a chord of a circle such that ∠BAC = 30°. The tangent at C intersects AB produced at D. Prove that BC = BD.
Solution
Join OC.
∠ ACB = 90° ...(Angle of the semicircle)
∠ ABC = 60° ...(Angle Sum property)
∠ CBD = 120° ...(adj to angle CBA 30°)
∠ OCD = 90° ...(tangent)
∠ COB = 60° ...(Angle at the center is equal to twice that of the circumference)
∠ OCB = 60° ...(Angle Sum property)
∠BCD = ∠ OCD - ∠OCB = 90° - 60° = 30°
∠BDC = ∠BCD = 30°
BD = BC
Hence proved.
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