18.4 In the convex quadrilateral $A B C D$, $A C$ bisects $\angle B A D$. A line $C E \perp A B$ is drawn through $C$ at $E$, and $A E=\frac{1}{2}(A B+A D)$. Then, the relationship between $\angle A B C$ and $\angle A D C$ is
(A) $\angle A B C=\angle A D C$.
(B) $\angle A B C+\angle A D C=180^{\circ}$.
(C) $\angle A B C+\angle A D C=150^{\circ}$.
(D) $\angle A B C \neq \angle A D C$.
(China Beijing Junior High School Mathematics Competition, 1991)
[Solution] $\because A E=\frac{1}{2}(A B+A D)$,
then
$$
A E=A B+A D-A E,
$$

or $A E=A D+E B$.
Extend $A D$ to $F$, such that $D F=E B$, then $A F=A E$.
Connect $F C$, and since $A C$ bisects $\angle B A D$,

then
$$
\triangle A E C \cong \triangle A F C,
$$

we have $\angle F=90^{\circ}$, and $C E=C F$.
Therefore, $\triangle C B E \cong \triangle C D F$, then $\angle B=\angle C D F, \angle D C F=\angle B C E$.
Thus $\angle A D C=\angle F+\angle D C F=\angle C E B+\angle B C E$.
$$
\therefore \quad \angle B+\angle A D C=\angle B+\angle C E B+\angle B C E=180^{\circ} \text {. }
$$

Hence, the answer is $(B)$.