Given regular pentagon $ABCDE$, a circle can be drawn that is tangent to $\overline{DC}$ at $D$ and to $\overline{AB}$ at $A$. The number of degrees in minor arc $AD$ is
Define major arc DA as $DA$, and minor arc DA as $da$. Extending DC and AB to meet at F, we see that $\angle CFB=36=\frac{DA-da}{2}$. We now have two equations: $DA-da=72$, and $DA+da=360$. Solving, $DA=216$ and $da=144\Rightarrow \mathrm{(E)}$.
Let $O$ be the center of the circle. Since the sum of the interior angles in any $n$-gon is $(n-2)180^\circ$, the sum of the angles in $ABCDO$ is $540^\circ$.
Since $\angle ABC=\angle BCD=108^\circ$ and $\angle OAB=\angle ODC= 90^{\circ}$, it follows that the measure of $\angle AOD$, and thus the measure of minor arc $AD$, equals $540^\circ - 108^\circ-108^\circ-90^\circ-90^\circ=\boxed{\mathrm{(E)}144^\circ}$.