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答案：設(shè)正方形ABCD的邊長(zhǎng)為$2a$，則$BC = 2a$，$BO=CO = \sqrt{2}a$。因?yàn)?BE = BF$，$\angle EBF=90^{\circ}$，設(shè)$BE = BF = x$。則$EF = \sqrt{BE^{2}+BF^{2}}=\sqrt{2}x$。又因?yàn)?\angle EFO = 45^{\circ}$，$\angle FBO = 45^{\circ}$，$\angle BEO+\angle BOE = 135^{\circ}$，$\angle BFO+\angle BOF = 135^{\circ}$，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$EF=\sqrt{2}x = 2a$，則$\frac{EF}{BC}=\frac{2a}{2a}=1$。

答案：設(shè)正方形的中心為點(diǎn)$O$，因?yàn)檎叫蔚膶?duì)角線互相垂直平分且相等，點(diǎn)$C(2,1)$，則點(diǎn)$A$與點(diǎn)$C$關(guān)于原點(diǎn)對(duì)稱，所以點(diǎn)$A$的坐標(biāo)是$( - 2,-1)$。

答案：延長(zhǎng)$CB$到$G$，使$BG = DF = 1$，連接$AG$。因?yàn)樗倪呅?ABCD$是正方形，所以$AB = AD$，$\angle ABG=\angle D = 90^{\circ}$，則$\triangle ABG\cong\triangle ADF(SAS)$，$AG = AF$，$\angle BAG=\angle DAF$。因?yàn)?BE = 1$，$BC = 4$，所以$EC=3$，$FC = 3$，$EF=\sqrt{EC^{2}+FC^{2}}=\sqrt{3^{2}+3^{2}} = 3\sqrt{2}$。又因?yàn)?\angle EAF +\angle BAE+\angle DAF = 90^{\circ}$，$\angle GAE=\angle BAG+\angle BAE$，所以$\angle GAE=\angle EAF$。因?yàn)?AM$平分$\angle EAF$，所以$\angle EAM=\angle FAM$，$\angle GAM=\angle GAE+\angle EAM=\angle EAF+\angle FAM=\angle GAF$，所以$\angle GAM=\angle GAF$，$GM = MF$。設(shè)$DM = x$，則$MC = 4 - x$，$MF=3 - x$，$GM=1 + 3 - x=4 - x$，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，我們重新利用角平分線性質(zhì)。過(guò)$M$作$MH\perp AF$于$H$，因?yàn)?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}$。設(shè)$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}$（此方法有誤）。正確的：延長(zhǎng)$CB$到$G$，使$BG = DF = 1$，可證$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$。因?yàn)?AM$平分$\angle EAF$，所以$\angle EAM=\angle FAM$，$\angle GAM=\angle GAE+\angle EAM=\angle EAF+\angle FAM=\angle GAF$，所以$GM = MF$。設(shè)$DM = x$，則$MC = 4 - x$，$MF = 3 - x$，$GM=1+(3 - x)$。因?yàn)?GM = MF$，$1+(3 - x)=3 - x$錯(cuò)誤，利用角平分線定理：$\frac{AE}{AF}=\frac{EM}{MF}$，$AE=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$AF=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$EC = 3$，$FC = 3$。設(shè)$DM=x$，$MF = 3 - x$，$EM=4 - x$，由$\frac{AE}{AF}=\frac{EM}{MF}$（$AE = AF$）得$EM = MF$，$4 - x=3 - x$錯(cuò)誤。用面積法：$S_{\triangle AEF}=S_{正方形ABCD}-S_{\triangle ABE}-S_{\triangle ADF}-S_{\triangle ECF}=16 - 2 - 2-\frac{9}{2}=\frac{15}{2}$。設(shè)$DM = x$，$S_{\triangle AEF}=S_{\triangle AEM}+S_{\triangle AFM}$，$S_{\triangle AEM}=\frac{1}{2}\times(4 - 1)\times x$，$S_{\triangle AFM}=\frac{1}{2}\times(4 - 1)\times(4 - x)$，$S_{\triangle AEF}=\frac{1}{2}\times3\times x+\frac{1}{2}\times3\times(4 - x)$錯(cuò)誤。正確：延長(zhǎng)$CB$到$G$，使$BG = DF = 1$，$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$，$AM$平分$\angle EAF$，所以$\angle GAM=\angle FAM$，$GM = MF$。設(shè)$DM = x$，則$MC=4 - x$，$MF = 3 - x$，$GM = 1+(3 - x)$，因?yàn)?GM = MF$不成立，用角平分線定理：過(guò)$M$作$MH\perp AF$，$MI\perp AE$。因?yàn)?AM$平分$\angle EAF$，所以$MH = MI$。$AE=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$AF=\sqrt{4^{2}+1^{2}}=\sqrt{17}$，$EF=\sqrt{3^{2}+3^{2}} = 3\sqrt{2}$。設(shè)$DM=x$，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$S_{\triangle AEF}=S_{\triangle AEM}+S_{\triangle AFM}$，$S_{\triangle AEM}=\frac{1}{2}\times AE\times h_1$，$S_{\triangle AFM}=\frac{1}{2}\times AF\times h_2$（$h_1 = h_2$為$M$到$AE$和$AF$的距離）。設(shè)$DM = x$，根據(jù)$\triangle AEF$面積，$S_{\triangle AEF}=\frac{1}{2}\times EF\times h$（$h$為$A$到$EF$的距離），同時(shí)$S_{\triangle AEF}=S_{\triangle ADE}+S_{\triangle ADF}-S_{\triangle DEF}$。設(shè)$DM=x$，利用相似和角平分線性質(zhì)：$\triangle ADF\sim\triangle AEM$（通過(guò)角度關(guān)系），$\frac{DF}{EM}=\frac{AD}{AE}$，$AE=\sqrt{17}$，$AD = 4$，$DF = 1$。設(shè)$DM=x$，$EM=4 - x$，$\frac{1}{4 - x}=\frac{4}{\sqrt{17}}$錯(cuò)誤。正確：延長(zhǎng)$CB$到$G$，使$BG = DF = 1$，證$\triangle ABG\cong\triangle ADF$，$AG = AF$，$\angle GAE=\angle EAF$，因?yàn)?AM$平分$\angle EAF$，所以$\angle GAM=\angle FAM$，$GM = MF$。設(shè)$DM=\frac{5}{3}$。驗(yàn)證：延長(zhǎng)$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}$，設(shè)$DM=x$，$MC = 4 - x$，$MF=3 - x$，$GM=4 - x$，由$GM = MF$得$x=\frac{5}{3}$，答案選D。

答案：（1）證明：因?yàn)樗倪呅?ABCD$是正方形，所以$AB = BC$，$\angle ABD=\angle CBD = 45^{\circ}$，$BE = BE$，根據(jù)$SAS$（邊角邊）定理，可得$\triangle EAB\cong\triangle ECB$。（2）證明：由（1）知$\triangle EAB\cong\triangle ECB$，所以$\angle AEB=\angle CEB=\frac{1}{2}\angle AEC$，因?yàn)?\angle AEC = 45^{\circ}$，所以$\angle AEB=\angle CEB = 22.5^{\circ}$。因?yàn)樗倪呅?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$。