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

It is given that ∠BAC = 30° and AB is diameter.
$\frac{\mathrm{AB}}{2}=\mathrm{OA}=\mathrm{OC}\left(\mathrm{Radius}\right)$

∠ACB = 90° (Angle formed by the diameter is 90°)

In ∆ABC,

∠ACB + ∠BAC + ∠ABC = 180°

⇒ 90° + 30° + ∠ABC = 180°

⇒ ∠ABC = 60°

⇒ ∠CBD = 180° – 60° = 120° ( ∠CBD and ∠ABC form a linear pair)

In ∆OCD,

∠OCD = 90° (Angle made by Radius on the tangent)

∠OBC = ∠ABC = 60°

Since OB = OC, ∠OCB = ∠OBC = 60° (OC = OB = radius)

In ∆OCB,

⇒ ∠COB + ∠OCB + ∠OBC = 180°

⇒ ∠COB + 60° + 60° = 180°

⇒ ∠COB = 60°

In ∆OCD,

∠COD + ∠OCD + ∠ODC = 180°

⇒ 60° + 90° + ∠ODC = 90° (∠COD = ∠COB)

⇒ ∠ODC = 30°

In ∆CBD,

∠CBD = 120°

∠BDC = ∠ODC = 30°

⇒ ∠BCD + ∠BDC + ∠CBD = 180°

⇒ ∠BCD + 30° + 120° = 180°

⇒ ∠BCD + 30° = ∠BDC

Angles made by BC and BD on CD are equal, so ∆CBD is an isosceles triangle and therefore, BC = BD.