(1) 求证: $\triangle A B F \cong \triangle E A D$;
(2) 如图 2. 若 $A B=9, C D=5, \angle E C F=\angle A E D$, 求 $B E$ 的长;
(3) 如图 3, 若 $B F$ 的延长线经过 $A D$ 的中点 $M$, 求 $\frac{B E}{E C}$ 的值.

\begin{aligned} &\therefore \angle A E B=\angle B C D, \\ & \angle A B C=\angle B C D, \\ &\therefore \angle A B C=\angle A E B, \\ &\therefore A B=A E, \\ & D E / / A B, \\ &\therefore \angle D E C=\angle A B C, \angle A E D=\angle B A F \\ & \angle A B C=\angle B C D, \\ &\therefore \angle D E C=\angle B C D, \\ &\therefore D E=D C, \\ & C F / / A D, A E / / C D, \\ &\therefore \text { 四边形 } A D C F \text { 是平行四边形, } \\ &\therefore A F=C D, \\ &\therefore A F=D E, \\ &\left\{\begin{array}{l} A B B F \cong A E A D=\text { 在 } \triangle A B F \text { 和 } \triangle E A D \text { 中, } \\ A F=D E \end{array}\right. \\ &\therefore \angle A E D, \\ &\therefore A B S) ; \end{aligned}

(2) 方法(1): $\mathrm{Q} C F / / A D$,
$\therefore \angle E A D=\angle C F E$ $\mathrm{Q} \angle E C F=\angle A E D$ $\therefore \triangle E A D C O \triangle C F E$ $\therefore \frac{A D}{E F}=\frac{D E}{C E}=\frac{A E}{C F}$

$$\therefore A D=C F, A F=C D \text {, }$$
$A B=9, C D=5$,
\begin{aligned} &\therefore A E=9, \quad D E=5, \\ &\therefore E F=A E-A F=9-5=4, \end{aligned}
$\therefore \frac{C F}{4}=\frac{5}{C E}=\frac{9}{C F}$,
$$\therefore C F^{2}=4 \times 9=36 \text {, 即 } C F=6 \text {, }$$
$\therefore C E=\frac{10}{3}$,
$$\angle A B C=\angle B C D=\angle A E B=\angle D E C \text {, }$$
$\therefore \triangle A B E^{\circ} \triangle D E C$,
$$\therefore \frac{B E}{A B}=\frac{E C}{D C} \text {, 即 } \frac{B E}{9}=\frac{\frac{10}{3}}{5} \text {, }$$

$$\therefore B E=6 \text {; }$$

\begin{aligned} &\therefore \angle A B F=\angle E A D, \\ & \angle E A D=\angle C F E \\ &\therefore \angle A B F=\angle C F E, \\ & \angle A B C=\angle A E B, \angle A B C=\angle A B F+\angle E B F, \angle A E B=\angle C F E+\angle E C F, \\ &\therefore \angle E B F=\angle E C F \\ & \angle B A E=\angle A E D=\angle E C F \\ &\therefore \angle E B F=\angle B A E \end{aligned}

\begin{aligned} & \angle B E F=\angle A E B, \\ &\therefore \triangle B E F \sim \triangle A E B, \\ &\therefore \frac{B E}{E F}=\frac{A E}{B E}, \text { 即 } \frac{B E}{4}=\frac{9}{B E}, \\ &\therefore B E=6 ; \end{aligned}
(3) 如图 3 , 延长 $B M 、 E D$ 交于点 $G$,
$\triangle A B E, \triangle D C E$ 均为等腰三角形, 且 $\angle A B C=\angle D C E$,
\begin{aligned} &\therefore \triangle A B E \backsim \triangle D C E, \\ &\therefore \frac{A B}{D C}=\frac{A E}{D E}=\frac{B E}{C E}, \end{aligned}

$$\therefore E F=A E-A F=a x-a=a(x-1) \text {, }$$
$A B / / D G$,
$$\therefore \angle A B G=\angle G$$
$A D$ 的中点 $M$,
$$\therefore A M=D M \text {, }$$
\begin{aligned} & \angle A M B=\angle D M G \\ &\therefore \triangle A M B \cong \triangle D M G(A A S) \end{aligned}

\begin{aligned} &\therefore D G=A B=a x, \\ &\therefore E G=D G+D E=a x+a=a(x+1), \\ &\therefore \frac{B E}{C E}=\frac{A B}{D E}=\frac{a x}{a}=x, \\ & A B / / D G \quad \text { 即 } A B / / E G), \\ &\therefore \triangle A B F C \Delta E G F, \\ &\therefore \frac{A B}{E G}=\frac{A F}{E F}, \text { 即 } \frac{a x}{a(x+1)}=\frac{a}{a(x-1)}, \\ &\therefore x^{2}-2 x-1=0, \end{aligned}

$$\therefore \frac{B E}{E C}=x=1+\sqrt{2} \text {. }$$

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