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.

Answer 1

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