如图, $\triangle A B C$ 中, 点 $D$ 是 $B C$ 延长线上一点, 且 $\angle C A D=90^{\circ}-\frac{1}{2} \angle B A C$, 过点 $C$ 作 $C E / / A D$ 交 $A B$ 于点 $E$,且 $\angle A C E=3 \angle B C E, A C=3, B E=2$, 则 $C D$ 的长为 $(\quad)$

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 $\because \angle C A D=90^{\circ}-\frac{1}{2} \angle B A C, \quad \therefore 2 \angle C A D=180^{\circ}-\angle B A C . \because \angle A E C+\angle A C E=180^{\circ}-\angle B A C$,

$\therefore 2 \angle C A D=\angle A E C+\angle A C E . \because C E / / A D, \quad \therefore \angle C A D=\angle A C E, \angle B C E=\angle B D A, \quad \therefore \angle A E C=\angle A C E$,

$\therefore A E=A C=3, \therefore A B=2+3=5 . \because \angle A C E=3 \angle B C E, \therefore \angle A E C=3 \angle B C E=\angle B+\angle B C E, \therefore 2 \angle B C E=$

$\angle B$. 如图, 作 $A F \perp C E$, 垂足为 $F$, 延长 $A F$ 交 $B C$ 于点 $M$, 作 $E G / / A F$, 交 $B C$ 于点 $G$, 连接 $E M . \because A E=A C$, $\therefore$ 易得 $A F$ 是 $C E$ 的垂直平分线, $\therefore C M=E M, \therefore \angle M C E=\angle M E C, \therefore \angle B M E=2 \angle M C E=\angle B, \therefore B E=$ $M E=M C=2$.

$\because E G / / A F, \quad \therefore \angle G E C=90^{\circ}, \frac{B E}{B G}=\frac{A E}{G M}, \quad \therefore M G=M E=M C=2, \quad \therefore \frac{2}{B G}=\frac{3}{2}, \quad \therefore B G=\frac{4}{3}$,

$\therefore B C=\frac{4}{3}+2+2=\frac{16}{3} . \because C E / / A D, \therefore \frac{B E}{B C}=\frac{A E}{C D}, \therefore \frac{2}{\frac{16}{3}}=\frac{3}{C D}$, 解得 $C D=8$. 故选 $A$.

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