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第49页

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 证明：

 (1)$\because$四边形$ABCD$为正方形，

 $\therefore AB = AD，$$\angle BAD=\angle B=\angle ADC = 90^{\circ}，$

 $\therefore \angle ADF = 90^{\circ}.$

 $\because \angle EAF = 90^{\circ}，$

 $\therefore \angle EAD+\angle FAD = 90^{\circ}，$$\angle EAD+\angle BAE = 90^{\circ}.$

 $\therefore \angle BAE=\angle DAF.$

 在$\triangle ABE$和$\triangle ADF$中，

$\begin{cases}\angle BAE=\angle DAF,\\AB = AD,\\\angle ABE=\angle ADF,\end{cases}$

 $\therefore \triangle ABE\cong\triangle ADF(\text{ASA}).$$\therefore BE = DF.$

 (2)$\because \triangle ABE\cong\triangle ADF，$$\therefore AE = AF.$

 $\because \angle EAF$的平分线交$CD$于点$G，$

 $\therefore \angle EAG=\angle FAG.$

 在$\triangle AEG$和$\triangle AFG$中，

$\begin{cases}AE = AF,\\\angle EAG=\angle FAG,\\AG = AG,\end{cases}$

 $\therefore \triangle AEG\cong\triangle AFG(\text{SAS}).$$\therefore GE = GF.$

 $\because GF = DG + DF，$而$BE = DF，$
$\therefore BE + DG = EG.$

 (1)证明：$\because$四边形$ABCD$是正方形，

 $\therefore \angle D=\angle A = 90^{\circ}.$

 $\because$四边形$EFGH$是菱形，$\therefore HG = HE.$

 $\because DG = AH = 2，$$\therefore \text{Rt}\triangle HDG\cong\text{Rt}\triangle EAH(\text{HL}).$

 $\therefore \angle DHG=\angle AEH.$$\therefore \angle DHG+\angle AHE = 90^{\circ}.$

 $\therefore \angle GHE = 90^{\circ}.$$\therefore$菱形$EFGH$为正方形.

 (2)解：如答图，过点$F$作$FM\perp CD，$交$DC$的延长线于点$M，$连接$GE.$

 $\because$四边形$ABCD$是正方形，$\therefore \angle A = 90^{\circ}，$$CD// AB.$

 $\therefore \angle AEG=\angle MGE.$

 $\because$四边形$EFGH$是菱形，$\therefore HE// GF，$$HE = GF.$

 $\therefore \angle HEG=\angle FGE.$$\therefore \angle HEA=\angle FGM.$

 $\because FM\perp CD，$$\therefore \angle M = 90^{\circ}.$$\therefore \angle M=\angle A.$

 在$\triangle AEH$和$\triangle MGF$中，

$\begin{cases}\angle A=\angle M,\\\angle AEH=\angle MGF,\\HE = FG,\end{cases}$

 $\therefore \triangle AEH\cong\triangle MGF(\text{AAS}).$$\therefore FM = AH = 2.$

 $\therefore S_{\triangle FCG}=\frac{1}{2}CG\cdot FM = 1.$

 $\therefore CG = 1.$$\therefore DG = DC - CG = 6 - 1 = 5.$