题目（八年级 · 画法几何学）：已知 $\triangle A B C$ 中, $A C=B C ; \triangle D E C$ 中, $D C=E C ; \angle A C B=\angle D C E=a$,



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图2

(1) 如图 1, 当 $a=60^{\circ}$ 时, (1)求证: $A D=B E$; (2)求出 $\angle A E B$ 的度数;

（2）如图 2, 当 $a=90^{\circ}$ 时, 求:

(1) $\angle A E B$ 的度数; (2)若 $\angle C A F=\angle B A F, B E=2$, 求 $A F$ 的长.
选项：
答案：解: (1) (1) $\because A C=B C, D C=E C, \angle A C B=\angle D C E=60^{\circ}$,

$\therefore \triangle A B C$ 和 $\triangle D E C$ 是等边三角形,

$\therefore \angle A C B=\angle D C E=\angle C D E=\angle C E D=60^{\circ}, C A=C B, C D=C E$,

$\therefore \angle A C B-\angle D C F=\angle D C E-\angle D C F$,

$\therefore \angle A C D=\angle B C E$,

在 $\triangle C D A$ 和 $\triangle C E B$ 中,

$\left\{\begin{aligned} A C & =B C \\ \angle A C D & =\angle B C E, \\ C D & =C E\end{aligned}\right.$

$\therefore \triangle C D A \cong \triangle C E B(S A S)$,

$\therefore A D=B E$

(2) $\because \triangle C D A \cong \triangle C E B$,

$\therefore \angle C E B=\angle C D A=180^{\circ}-\angle C D E=120^{\circ}$,

$\because \angle C E D=60^{\circ}$,

$\therefore \angle A E B=\angle C E B-\angle C E D=120^{\circ}-60^{\circ}=60^{\circ}$;

(2) (1) $\because A C=B C, C D=C E, \angle A C B=\angle D C E=90^{\circ}$,

$\therefore \triangle A C B$ 和 $\triangle D C E$ 均为等腰直角三角形,

$\therefore \angle C D E=45^{\circ}=\angle C E D$,

$\therefore \angle A C B-\angle D C B=\angle D C E-\angle D C B, \quad$ 即 $\angle A C D=\angle B C E$,

$\therefore \angle A D C=180^{\circ}-\angle C D E=135^{\circ}$,

在 $\triangle A C D$ 和 $\triangle B C E$ 中,

$\left\{\begin{aligned} A C & =B C \\ \angle A C D & =\angle B C E, \\ C D & =C E\end{aligned}\right.$

$\therefore \triangle A C D \xlongequal{\triangle} \triangle B C E(S A S)$,

$\therefore \angle A D C=\angle B E C=135^{\circ}$,

$\therefore \angle A E B=\angle B E C-\angle C E D=90^{\circ}$,

(2) $\because \triangle A C D \xlongequal{\triangle} \triangle B C E, B E=2$,

$\therefore B E=A D=2$,

$\because \angle C A F=\angle B A F=22.5^{\circ}, \angle C D E=45^{\circ}=\angle C A D+\angle A C D$,

$\therefore \angle A C D=\angle C A D=22.5^{\circ}$,

$\therefore A D=C D=2$,

$\because \angle D C F=90^{\circ}-\angle A C D=67.5^{\circ}, \angle A F C=\angle A B C+\angle B A F=67.5^{\circ}$,

$\therefore \angle D C F=\angle A F C$

$\therefore D C=D F=2$,

$\therefore A F=A D+D F=4$,
解析：解: (1) (1) $\because A C=B C, D C=E C, \angle A C B=\angle D C E=60^{\circ}$,

$\therefore \triangle A B C$ 和 $\triangle D E C$ 是等边三角形,

$\therefore \angle A C B=\angle D C E=\angle C D E=\angle C E D=60^{\circ}, C A=C B, C D=C E$,

$\therefore \angle A C B-\angle D C F=\angle D C E-\angle D C F$,

$\therefore \angle A C D=\angle B C E$,

在 $\triangle C D A$ 和 $\triangle C E B$ 中,

$\left\{\begin{aligned} A C & =B C \\ \angle A C D & =\angle B C E, \\ C D & =C E\end{aligned}\right.$

$\therefore \triangle C D A \cong \triangle C E B(S A S)$,

$\therefore A D=B E$

(2) $\because \triangle C D A \cong \triangle C E B$,

$\therefore \angle C E B=\angle C D A=180^{\circ}-\angle C D E=120^{\circ}$,

$\because \angle C E D=60^{\circ}$,

$\therefore \angle A E B=\angle C E B-\angle C E D=120^{\circ}-60^{\circ}=60^{\circ}$;

(2) (1) $\because A C=B C, C D=C E, \angle A C B=\angle D C E=90^{\circ}$,

$\therefore \triangle A C B$ 和 $\triangle D C E$ 均为等腰直角三角形,

$\therefore \angle C D E=45^{\circ}=\angle C E D$,

$\therefore \angle A C B-\angle D C B=\angle D C E-\angle D C B, \quad$ 即 $\angle A C D=\angle B C E$,

$\therefore \angle A D C=180^{\circ}-\angle C D E=135^{\circ}$,

在 $\triangle A C D$ 和 $\triangle B C E$ 中,

$\left\{\begin{aligned} A C & =B C \\ \angle A C D & =\angle B C E, \\ C D & =C E\end{aligned}\right.$

$\therefore \triangle A C D \xlongequal{\triangle} \triangle B C E(S A S)$,

$\therefore \angle A D C=\angle B E C=135^{\circ}$,

$\therefore \angle A E B=\angle B E C-\angle C E D=90^{\circ}$,

(2) $\because \triangle A C D \xlongequal{\triangle} \triangle B C E, B E=2$,

$\therefore B E=A D=2$,

$\because \angle C A F=\angle B A F=22.5^{\circ}, \angle C D E=45^{\circ}=\angle C A D+\angle A C D$,

$\therefore \angle A C D=\angle C A D=22.5^{\circ}$,

$\therefore A D=C D=2$,

$\because \angle D C F=90^{\circ}-\angle A C D=67.5^{\circ}, \angle A F C=\angle A B C+\angle B A F=67.5^{\circ}$,

$\therefore \angle D C F=\angle A F C$

$\therefore D C=D F=2$,

$\therefore A F=A D+D F=4$,