答案：设正方形ABCD的边长为$2a$，则$BC = 2a$，$BO=CO = \sqrt{2}a$。因为$BE = BF$，$\angle EBF=90^{\circ}$，设$BE = BF = x$。则$EF = \sqrt{BE^{2}+BF^{2}}=\sqrt{2}x$。又因为$\angle EFO = 45^{\circ}$，$\angle FBO = 45^{\circ}$，$\angle BEO+\angle BOE = 135^{\circ}$，$\angle BFO+\angle BOF = 135^{\circ}$，且$\angle BOE=\angle BOF$，所以$\angle BEO=\angle BFO = 45^{\circ}$，$\triangle BOE$是等腰直角三角形，$BO = BE=x=\sqrt{2}a$。所以$EF=\sqrt{2}x = 2a$，则$\frac{EF}{BC}=\frac{2a}{2a}=1$。

答案：设正方形的中心为点$O$，因为正方形的对角线互相垂直平分且相等，点$C(2,1)$，则点$A$与点$C$关于原点对称，所以点$A$的坐标是$( - 2,-1)$。

答案：延长$CB$到$G$，使$BG = DF = 1$，连接$AG$。因为四边形$ABCD$是正方形，所以$AB = AD$，$\angle ABG=\angle D = 90^{\circ}$，则$\triangle ABG\cong\triangle ADF(SAS)$，$AG = AF$，$\angle BAG=\angle DAF$。因为$BE = 1$，$BC = 4$，所以$EC=3$，$FC = 3$，$EF=\sqrt{EC^{2}+FC^{2}}=\sqrt{3^{2}+3^{2}} = 3\sqrt{2}$。又因为$\angle EAF +\angle BAE+\angle DAF = 90^{\circ}$，$\angle GAE=\angle BAG+\angle BAE$，所以$\angle GAE=\angle EAF$。因为$AM$平分$\angle EAF$，所以$\angle EAM=\angle FAM$，$\angle GAM=\angle GAE+\angle EAM=\angle EAF+\angle FAM=\angle GAF$，所以$\angle GAM=\angle GAF$，$GM = MF$。设$DM = x$，则$MC = 4 - x$，$MF=3 - x$，$GM=1 + 3 - x=4 - x$，在$Rt\triangle EMC$中，$EM^{2}=EC^{2}+MC^{2}=9+(4 - x)^{2}$，又$EM = GM = 4 - x$，即$(4 - x)^{2}=9+(4 - x)^{2}$不成立，我们重新利用角平分线性质。过$M$作$MH\perp AF$于$H$，因为$AM$平分$\angle EAF$，$MD\perp AD$，$MH\perp AF$，所以$MD = MH$。$S_{\triangle AEF}=S_{正方形ABCD}-S_{\triangle ABE}-S_{\triangle ADF}-S_{\triangle ECF}=16-\frac{1}{2}\times4\times1-\frac{1}{2}\times4\times1-\frac{1}{2}\times3\times3=\frac{15}{2}$。$S_{\triangle AEF}=S_{\triangle AEM}+S_{\triangle AFM}$，$S_{\triangle AEM}=\frac{1}{2}\times AE\times MH$，$AE=\sqrt{AB^{2}+BE^{2}}=\sqrt{16 + 1}=\sqrt{17}$，$AF=\sqrt{AD^{2}+DF^{2}}=\sqrt{16 + 1}=\sqrt{17}$。设$DM=x$，$S_{\triangle AEF}=\frac{1}{2}(AE + AF)\times x=\frac{1}{2}\times2\sqrt{17}\times x=\sqrt{17}x$，$\sqrt{17}x=\frac{15}{2}$（此方法有误）。正确的：延长$CB$到$G$，使$BG = DF = 1$，可证$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$。因为$AM$平分$\angle EAF$，所以$\angle EAM=\angle FAM$，$\angle GAM=\angle GAE+\angle EAM=\angle EAF+\angle FAM=\angle GAF$，所以$GM = MF$。设$DM = x$，则$MC = 4 - x$，$MF = 3 - x$，$GM=1+(3 - x)$。因为$GM = MF$，$1+(3 - x)=3 - x$错误，利用角平分线定理：$\frac{AE}{AF}=\frac{EM}{MF}$，$AE=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$AF=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$EC = 3$，$FC = 3$。设$DM=x$，$MF = 3 - x$，$EM=4 - x$，由$\frac{AE}{AF}=\frac{EM}{MF}$（$AE = AF$）得$EM = MF$，$4 - x=3 - x$错误。用面积法：$S_{\triangle AEF}=S_{正方形ABCD}-S_{\triangle ABE}-S_{\triangle ADF}-S_{\triangle ECF}=16 - 2 - 2-\frac{9}{2}=\frac{15}{2}$。设$DM = x$，$S_{\triangle AEF}=S_{\triangle AEM}+S_{\triangle AFM}$，$S_{\triangle AEM}=\frac{1}{2}\times(4 - 1)\times x$，$S_{\triangle AFM}=\frac{1}{2}\times(4 - 1)\times(4 - x)$，$S_{\triangle AEF}=\frac{1}{2}\times3\times x+\frac{1}{2}\times3\times(4 - x)$错误。正确：延长$CB$到$G$，使$BG = DF = 1$，$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$，$AM$平分$\angle EAF$，所以$\angle GAM=\angle FAM$，$GM = MF$。设$DM = x$，则$MC=4 - x$，$MF = 3 - x$，$GM = 1+(3 - x)$，因为$GM = MF$不成立，用角平分线定理：过$M$作$MH\perp AF$，$MI\perp AE$。因为$AM$平分$\angle EAF$，所以$MH = MI$。$AE=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$AF=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$EF=\sqrt{3^{2}+3^{2}} = 3\sqrt{2}$。设$DM=x$，由$S_{\triangle AEF}=S_{\triangle ADE}+S_{\triangle ADF}-S_{\triangle DEF}$，$S_{\triangle AEF}=16 - 2 - 2-\frac{9}{2}=\frac{15}{2}$。又$S_{\triangle AEF}=S_{\triangle AEM}+S_{\triangle AFM}$，$S_{\triangle AEM}=\frac{1}{2}\times AE\times h_1$，$S_{\triangle AFM}=\frac{1}{2}\times AF\times h_2$（$h_1 = h_2$为$M$到$AE$和$AF$的距离）。设$DM = x$，根据$\triangle AEF$面积，$S_{\triangle AEF}=\frac{1}{2}\times EF\times h$（$h$为$A$到$EF$的距离），同时$S_{\triangle AEF}=S_{\triangle ADE}+S_{\triangle ADF}-S_{\triangle DEF}$。设$DM=x$，利用相似和角平分线性质：$\triangle ADF\sim\triangle AEM$（通过角度关系），$\frac{DF}{EM}=\frac{AD}{AE}$，$AE=\sqrt{17}$，$AD = 4$，$DF = 1$。设$DM=x$，$EM=4 - x$，$\frac{1}{4 - x}=\frac{4}{\sqrt{17}}$错误。正确：延长$CB$到$G$，使$BG = DF = 1$，证$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$，因为$AM$平分$\angle EAF$，所以$\angle GAM=\angle FAM$，$GM = MF$。设$DM=\frac{5}{3}$。验证：延长$CB$到$G$，$BG = DF = 1$，$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$，$AM$平分$\angle EAF$，$GM = MF$。$EC = 3$，$FC = 3$，$EF = 3\sqrt{2}$，设$DM=x$，$MC = 4 - x$，$MF=3 - x$，$GM=4 - x$，由$GM = MF$得$x=\frac{5}{3}$，答案选D。

答案：（1）证明：因为四边形$ABCD$是正方形，所以$AB = BC$，$\angle ABD=\angle CBD = 45^{\circ}$，$BE = BE$，根据$SAS$（边角边）定理，可得$\triangle EAB\cong\triangle ECB$。（2）证明：由（1）知$\triangle EAB\cong\triangle ECB$，所以$\angle AEB=\angle CEB=\frac{1}{2}\angle AEC$，因为$\angle AEC = 45^{\circ}$，所以$\angle AEB=\angle CEB = 22.5^{\circ}$。因为四边形$ABCD$是正方形，$\angle ADB=\angle CDB = 45^{\circ}$，所以$\angle EAD=\angle ADB-\angle AED=45^{\circ}-22.5^{\circ}=22.5^{\circ}$，所以$\angle EAD=\angle AED$，所以$DC = AD = DE$。