Let $O$ be the circumcentre of $\triangle ABC$ and $E$ the point where the line $AD$ meets the circumcircle again.

Because central angles are twice inscribed angles subtended by the same arc, we obtain $\angle COA = 2\angle CBA = 60^\circ$ and $\angle EOC = 2\angle EAC = 60^\circ$. Using this and $|OA| = |OC| = |OE|$, we see that the triangles $OAC$ and $OCE$ are equilateral. This implies that $AE$ is the perpendicular bisector of $OC$ and we obtain $|DO| = |DC|$ and $|EO| = |EC|$. This implies that $\angle DOE$ is congruent to $\angle DCE$, in particular, $\angle ODE = \angle CDE = \angle ADB = 45^\circ$ and hence $OD$ is perpendicular to $BC$. This implies that $D$ is the mid point of $BC$, as required.

In triangle $ABD$ we have $\angle BAD = 180^\circ - 30^\circ - 45^\circ = 105^\circ$. Because $\sin 105^\circ = \cos 15^\circ$ and $\sin 30^\circ = 2 \sin 15^\circ \cos 15^\circ$, with the aid of the sine rule for $\triangle ABD$ we obtain
$$|BD| = \frac{\sin 105^\circ}{\sin 30^\circ} |AD| = \frac{1}{2 \sin 15^\circ} |AD|.$$
On the other hand, the sine rule for $\triangle ADC$ gives
$$|DC| = \frac{\sin 30^\circ}{\sin 15^\circ} |AD| = 2 \cos 15^\circ |AD|.$$
Finally, using $\sin 30^\circ = 1/2$ we get $1 = 2 \sin 30^\circ = 4 \sin 15^\circ \cos 15^\circ$ and this implies
$$$$
$\frac{1}{2 \sin 15^\circ}$ = 2 $\cos 15^\circ, \quad \text{hence} \quad$ |BD| = |DC|.