jmc

Let $O$ be the center of the circle. The sum of the angles in pentagon $ABCDO$ is $3(180^{\circ })=540^{\circ }.$ Since $\angle ABC$ and $\angle BCD$ are interior angles of a regular pentagon, they each measure $108^{\circ }.$ The given circle is tangent to $\overline{AB}$ at $A$ and to $\overline{CD}$ at $D,$ and so it follows that $\angle OAB=\angle ODC=90^{\circ }.$ Then $$\begin{array}{rl}\angle AOD& =540^{\circ }-\angle ABC-\angle BCD-\angle OAB-\angle ODC\\ & =540^{\circ }-2(108^{\circ })-2(90^{\circ })=144^{\circ }.\end{array}$$Thus, the measure of minor arc $AD$ is also $\boxed{144^{\circ }}.$