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 extended AB at a point D. Prove that BC = BD.

Answer 1

Given, AB is a diameter and AC is a chord of circle with centre O, ∠BAC = 30°

To prove: BC = BD

Join BC

∠BCD = ∠CAB  ......[Angles in alternate segment]

∠CAB = 30° .....[Given]

∠BCD = 30° ......(i)

∠ACB = 90° .......[Angle in semi-circle]

In ∆ABC,

∠A + ∠B + ∠C = 180°  .......[Angle sum property]

30° + ∠CBA + 90° = 180°

⇒ ∠CBA = 60°

Also, ∠CBA + ∠CBD = 180°  ........[Linear pair]

⇒ ∠CBD = 180° – 60° = 120°    .......[∵ ∠ CBA = 60°]

Now, in ACBD,

∠CBD + ∠BDC + ∠DCB = 180°

⇒ 120° + ∠BDC + 30° = 180°

⇒ ∠BDC = 30°  ........(ii)

From (i) and (ii),

∠BCD = ∠BDC

⇒ BC = BD  .......[Sides opposite to equal angles are equal]