Write 'True' or 'False' and give reasons for your answer.
$AB$ is a diameter of a circle and $AC$ is its chord such that $\angle BAC = 30^{\circ}$. If the tangent at $C$ intersects $AB$ extended at $D$,then $BC = BD$.

(A) True
To prove: $BC = BD$.
$1$. Join $OC$. Since $AB$ is the diameter,$\angle ACB = 90^{\circ}$ (angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Since $OC$ is the radius and $CD$ is the tangent,$OC \perp CD$,so $\angle OCD = 90^{\circ}$.
$4$. In $\triangle OAC$,$OA = OC$ (radii),so $\angle OCA = \angle OAC = 30^{\circ}$.
$5$. $\angle BCD = \angle OCD - \angle OCB = 90^{\circ} - (90^{\circ} - 30^{\circ}) = 30^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$8$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the sides opposite to these angles are equal,hence $BC = BD$.