题目（八年级 · 画法几何学）：如图, 在平行四边形 $A B C D$ 中, $\angle C=60^{\circ}$, 过点 $D$ 作 $D E \perp B C$ 交 $B C$ 于点 $E$, 且 $D E=A D, F$ 为 $D C$ 上一点, 且 $A D=F D$, 连接 $A F$ 与 $D E$ 交于点 $G$.

(1)求证: $G F=G D=C E$.

(2)过点 $A$ 作 $A H \perp A D$, 且 $A H=C E$, 求证: $A B=D G+A H$.

选项：
答案：(1)证明: $\because$ 四边形 $A B D$ 是平行四边形,
$\therefore \angle B A D=\angle C=60^{\circ}, A D / / B C, A B / / C D$,

$\therefore \angle A D C+\angle C=180^{\circ}, \angle A D G=\angle D E C$,

$\therefore \angle A D C=120^{\circ}$,

$\because D E \perp B C$

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

$\therefore \angle G D F=30^{\circ}$,

$\because A D=F D$,

$\therefore \angle D A F=\angle G D F=30^{\circ}$,

$\therefore \angle B A F=\angle B A D-\angle D A F=30^{\circ}$,

$\because A B / / C D$

$\therefore \angle G F D=\angle B A F=30^{\circ}=\angle G D F$,

$\therefore G F=G D$,

在 $\triangle A D G$ 和 $\triangle D E C$ 中, $\left\{\begin{array}{l}\angle A D G=\angle D E C \\ A D=D E \\ \angle D A G=\angle E D C\end{array}\right.$,

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

$\therefore G D=C E$,

$\therefore G F=G D=C E$

(2)证明: $\because A H \perp A D, D E \perp B C$,

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

在 $\triangle D A H$ 与 $\triangle D E C$ 中, $\left\{\begin{array}{l}A H=C E \\ \angle D A H=\angle D E C \\ A D=D E\end{array}\right.$,

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

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

$\because$ 四边形 $A B C D$ 是平行四边形,

$\therefore A B=C D, A B / / C D$,

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

$\because A D=D F$,

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

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$\therefore \angle D F A=\frac{1}{2} \angle C$,
在 $D H$ 上截取 $H M=A H$, 如图所示:

$\therefore \angle H A M=\angle H M A$,

$\therefore \angle H=180^{\circ}-2 \angle H A M$

$\because \angle M A D=90^{\circ}-\angle H A M$

$\therefore \angle D A M=\frac{1}{2} \angle H$,

$\therefore \angle M A D=\angle G F D$

在 $\triangle A D M$ 与 $\triangle F D G$ 中, $\left\{\begin{array}{l}\angle M A D=\angle G F D \\ A D=F D \\ \angle A D M=\angle F D G\end{array}\right.$,

$\therefore \triangle A D M \cong \triangle F D G(A S A)$,

$\therefore D M=D G$,

$\because A B=C D=D H=H M+D M$,

$\therefore A B=A H+D G$.
解析：(1)证明: $\because$ 四边形 $A B D$ 是平行四边形,
$\therefore \angle B A D=\angle C=60^{\circ}, A D / / B C, A B / / C D$,

$\therefore \angle A D C+\angle C=180^{\circ}, \angle A D G=\angle D E C$,

$\therefore \angle A D C=120^{\circ}$,

$\because D E \perp B C$

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

$\therefore \angle G D F=30^{\circ}$,

$\because A D=F D$,

$\therefore \angle D A F=\angle G D F=30^{\circ}$,

$\therefore \angle B A F=\angle B A D-\angle D A F=30^{\circ}$,

$\because A B / / C D$

$\therefore \angle G F D=\angle B A F=30^{\circ}=\angle G D F$,

$\therefore G F=G D$,

在 $\triangle A D G$ 和 $\triangle D E C$ 中, $\left\{\begin{array}{l}\angle A D G=\angle D E C \\ A D=D E \\ \angle D A G=\angle E D C\end{array}\right.$,

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

$\therefore G D=C E$,

$\therefore G F=G D=C E$

(2)证明: $\because A H \perp A D, D E \perp B C$,

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

在 $\triangle D A H$ 与 $\triangle D E C$ 中, $\left\{\begin{array}{l}A H=C E \\ \angle D A H=\angle D E C \\ A D=D E\end{array}\right.$,

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

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

$\because$ 四边形 $A B C D$ 是平行四边形,

$\therefore A B=C D, A B / / C D$,

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

$\because A D=D F$,

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



$\therefore \angle D F A=\frac{1}{2} \angle C$,
在 $D H$ 上截取 $H M=A H$, 如图所示:

$\therefore \angle H A M=\angle H M A$,

$\therefore \angle H=180^{\circ}-2 \angle H A M$

$\because \angle M A D=90^{\circ}-\angle H A M$

$\therefore \angle D A M=\frac{1}{2} \angle H$,

$\therefore \angle M A D=\angle G F D$

在 $\triangle A D M$ 与 $\triangle F D G$ 中, $\left\{\begin{array}{l}\angle M A D=\angle G F D \\ A D=F D \\ \angle A D M=\angle F D G\end{array}\right.$,

$\therefore \triangle A D M \cong \triangle F D G(A S A)$,

$\therefore D M=D G$,

$\because A B=C D=D H=H M+D M$,

$\therefore A B=A H+D G$.