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.

Answer 1

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.