Let the incircle $W$ with radius $r=(AC+CB-AB)/2$ of $\triangle ABC$ intersect $AB$ at $G$. $W$ is tangent to $OD$ if and only if $OG=r\Leftrightarrow AB/2-(AB+BC-AC)/2=(AC+CB-AB)/2\Leftrightarrow AB=2BC\Leftrightarrow \angle CAB=30$

Letting $OA=OB=R, \angle CAB=\theta \Longrightarrow CD=2R\cos\theta-R\sec \theta$, $OD=R\tan \theta, BC=2R\sin\theta$. Now, $OB+CD=OD+BC\Longrightarrow 1+2\cos\theta-\sec\theta=\tan\theta+2\sin\theta$. Upon simplification, this gives $\sin 2\theta=\sin 4\theta$. Only possible value of $\theta$ is $\frac{\pi}{6}$.

Another easy way for the proof: Say that the center of the tangential quadrilateral is $W$ then: First prove that $D,W,B$ are linear then prove that $\frac{\angle(CAB)}{2}+45^{\circ}=\angle{OBW}=\alpha$ after that use tangents \[tan(\angle{ODB})=\frac{OB}{OD}=tan(2\alpha)\] \[tan\angle{ADO}=\frac{OD}{AO}=\frac{OD}{OB}=tan(45^{\circ}+\alpha)\Longrightarrow tan(2\alpha)=\frac{1}{tan(45^{\circ}+\alpha)}\]\[\Longleftrightarrow \] \[cot(2\alpha)=tan(45^{\circ}+\alpha)\] and this eqaution has got one solution for this question $45^{\circ}+\alpha+2\alpha=90^{\circ} \Longleftrightarrow \alpha=15=\frac{\angle(CAB)}{2} \Longrightarrow \boxed{\angle(CAB)=30^{\circ}}$