18.103 As shown in the figure, if line segments $AB$ and $CD$ are parallel, and $\angle D = 2 \angle B$, $AD = a$, $CD = b$, then the length of $AB$ is 
(A) $\frac{1}{2} a + 2 b$.
(B) $\frac{3}{2} b + \frac{3}{4} a$.
(C) $2 a - b$.
(D) $4 b - \frac{1}{2} a$.
(E) $a + b$.
(21st American High School Mathematics Examination, 1970)
［Solution］Draw the angle bisector $DE$ of $\angle D$ intersecting $AB$ at $E$.
Given $AB \parallel CD$, and $\angle D = 2 \angle B$,
thus $\angle CDE = \angle B$, and $ED \parallel BC$.
$\therefore BE = DC = b$.
Also, $\angle AED = \angle B = \angle CDE = \angle ADE$,

then
$$
\begin{array}{l} 
AE = AD = a . \\
\therefore \quad AB = AE + BE = a + b . \\
\text { Therefore, the answer is }(E) .
\end{array}
$$

Therefore, the answer is $(E)$.