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

(N/A) $1$. In $\triangle ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Let $OC$ be the radius. Since $DC$ is a tangent,$OC \perp CD$. Thus,$\angle OCD = 90^{\circ}$.
$4$. $\angle OCA = \angle OAC = 30^{\circ}$ (Since $OA = OC$,$\triangle OAC$ is isosceles).
$5$. $\angle BCD = \angle OCD - \angle OCA = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (Linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 60^{\circ}) = 0^{\circ}$? Wait,let us re-evaluate: $\angle BCD = 60^{\circ}$,$\angle CBD = 120^{\circ}$,then $\angle BDC = 180^{\circ} - 180^{\circ} = 0^{\circ}$ is incorrect. Let's use the property: $\angle BCD = \angle BAC = 30^{\circ}$ (Angle between tangent and chord equals angle in alternate segment).
$8$. $\angle BCD = 30^{\circ}$. In $\triangle BCD$,$\angle CBD = 120^{\circ}$,$\angle BCD = 30^{\circ}$,so $\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$9$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the triangle $BCD$ is isosceles with $BC = BD$.