smc

Because $△ACD$ is isosceles with $AC=CD$, $\measuredangle CAD=\measuredangle CDA=\theta$, where $\theta$ is some angle measure. Because $\angle CDA$ and $\angle ADB$ form a straight angle, $\measuredangle ADB=180^{\circ }-\theta$. Thus, because the interior angles of a triangle add to $180^{\circ }$, $\measuredangle BAD+\measuredangle ABC=\theta$, so $\measuredangle ABC=\theta -\measuredangle BAD$. Notice that $\measuredangle CAB=\theta +\measuredangle BAD$. With all of this information and recalling that, from the problem, $\measuredangle CAB-\measuredangle ABC=30^{\circ }$, we see that: $$\begin{array}{rl}\measuredangle CAB-\measuredangle ABC& =30^{\circ }\\ \theta +\measuredangle BAD-(\theta -\measuredangle BAD)& =30^{\circ }\\ 2\measuredangle BAD& =30^{\circ }\\ \measuredangle BAD& =15^{\circ }\end{array}$$ Thus, our answer is $\boxed{15^{\circ }}$.