20.98 Given a circle that is tangent to side $DC$ of a regular pentagon $ABCDE$ at $D$, and to side $AB$ at $A$, the measure of the minor arc $AD$ is
(A) 72.
(B) 108.
(C) 120.
(D) 135.
(E) 144.
(46th American High School Mathematics Examination, 1995)
[Solution] Connect $A D$. Given that $A B C D E$ is a regular pentagon, we know $\angle E=108^{\circ}$.
Also, $\triangle E A D$ is an isosceles triangle, so
$$
\begin{aligned}
\angle E A D & =\left(180^{\circ}-108^{\circ}\right) \div 2=36^{\circ}, \\
\therefore \quad \angle D A B & =108^{\circ}-36^{\circ}=72^{\circ},
\end{aligned}
$$

By the inscribed angle theorem, we have $\overparen{A D} \stackrel{\mathrm{m}}{=} 144^{\circ}$.
Therefore, the answer is $(E)$