In the figure,$O$ is the centre of the circle,$BD = OD$ and $CD \perp AB$. Find $\angle CAB$.

$(30^{\circ})$ In $\Delta OBD$,we have $OB = OD$ (radii of the same circle) and $BD = OD$ (given).
Since $OB = OD = BD$,$\Delta OBD$ is an equilateral triangle.
Therefore,$\angle BOD = 60^{\circ}$.
In $\Delta ODC$ and $\Delta BDC$,we have $OD = BD$ (given),$\angle ODC = \angle BDC$ (since $\Delta ODC \cong \Delta BDC$ by $SAS$ as $CD$ is common and $OC=OB$),or more simply,since $CD \perp AB$,$CD$ is the perpendicular bisector of $OB$ if $D$ were on $AB$,but here $CD$ is a chord perpendicular to diameter $AB$ at point $P$. Let $CD$ intersect $AB$ at $P$.
In $\Delta OPD$ and $\Delta BPD$,$OD = BD$ (given),$DP = DP$ (common),$\angle OPD = \angle BPD = 90^{\circ}$.
Thus,$\Delta OPD \cong \Delta BPD$ by $RHS$ congruence rule.
This implies $\angle DOP = \angle DBP$. Since $\angle BOD = 60^{\circ}$,$\angle DOP = 60^{\circ}$.
Thus,$\angle DBP = 60^{\circ}$.
In $\Delta BPD$,$\angle BDP = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Since angles subtended by the same arc $CD$ at the circumference are equal,$\angle CAB = \angle CDB$.
Therefore,$\angle CAB = 30^{\circ}$.