Let the line $AD$ intersect the circumcircle of the triangle $ABC$ a second time at point $D'$; by the assumptions, $\angle CDD' = \angle D'DB = \angle CAB$ (see figure below).

We show that $AD = DD'$. As the quadrilateral $ABD'C$ is cyclic, $\angle ABC = \angle AD'C = \angle DD'C$ and $\angle BCA = \angle BD'A = \angle BD'D$. Thus the triangles $ABC$, $DD'C$ and $DBD'$ are similar. Hence $\frac{|DD'|}{|DB|} = \frac{|DC|}{|DD'|}$, implying $|DD'|^2 = |BD| \cdot |CD|$.

By similarity of the triangles $ABC$ and $DD'C$, we obtain $\angle BCA = \angle D'CD$, which implies $\angle DCA = \angle D'CB = \angle D'AB = \angle DAB$. By similarity of the triangles $ABC$ and $DBD'$ we analogously get $\angle ABD = \angle CBD' = \angle CAD' = \angle CAD$. Hence the triangles $ABD$ and $CAD$ are similar. Consequently, $\frac{AD}{CD} = \frac{BD}{AD}$, which implies $AD^2 = BD \cdot CD$.

Altogether, we have proven $AD^2 = DD'^2$, which implies the desired result.

Let the line $AD$ intersect the circumcircle of the triangle $ABC$ a second time at point $D'$; by the assumptions, $\angle CDD' = \angle D'DB = \angle CAB$. We show that $AD = DD'$.

Let $O$ be the circumcenter of the triangle $ABC$. Let the line $AD$ intersect the circumcircle of the triangle $BCD$ a second time at point $A'$ (see figure below).

By the equality $\angle CDA' = \angle A'DB$, the arcs $BA'$ and $A'C$ of the circumcircle of the triangle $BCD$ are equal.

By the conditions of the problem, $\angle CDB = 2\angle CAB$. Hence point $O$ lies on the circumcircle of the triangle $BCD$. Since $OC = OB$, the arcs $CO$ and $OB$ of the circumcircle of the triangle $BCD$ are equal.

Consequently, $OA'$ is a diameter of the circumcircle of the triangle $BCD$. If $O \neq D$ then, by Thales' theorem, $\angle A'DO = 90^\circ$. Thus $OD \perp AD'$, which implies $AD = DD'$ since a radius perpendicular to a chord bisects the chord. If $O = D$ (see figure below) then $AD'$ is a diameter of the circumcircle of the triangle $ABC$, obviously bisected by the center $O$.