Question

AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD.

Answer 1

Required to prove: BC = BD

Join BC and OC.

Given, ∠BAC = 30°

⇒ ∠BCD = 30°

[angle between tangent and chord is equal to angle made by chord in the alternate segment]

∠ACD = ∠ACO + ∠OCD

∠ACD = 30° + 90° = 120°

[OC ⊥ CD and OA = OC = radius ⇒ ∠OAC = ∠OCA = 30°]

In ∆ACD,

∠CAD + ∠ACD + ∠ADC = 180° [Angle sum property of a triangle]

⇒ 30° + 120° + ∠ADC = 180°

⇒ ∠ADC = 180° – (30° + 120°) = 30°

Now, in ∆BCD,

∠BCD = ∠BDC = 30°

⇒ BC = BD [As sides opposite to equal angles are equal]

• Hence Proved