题目（八年级 · 画法几何学）：如图,正方形$ABCD$中,$E$是对角线$BD$上一点,连接$AE$,过点$E$作$EFperpAE$,交边$BC$于点$F$.(1)求证:$EA=EF$;(2)写出线段$FC,DE$的数量关系并加以证明;(3)若$AB=4,FE=FC$,求$DE$的长.
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答案：(1)过点$E$作$MNperpAD$于$M$,交$BC$于点$N$,由四边形$ABCD$为正方形,$AEperpEF$,可证明$triangleAEMcongtriangleEFN(ASA)$，即可得$AE=EF$.(2)由$triangleAEMcongtriangleEFN,angleADB=45^{circ}$，可得$CF=2MD$,而$DE=sqrt{2}MD$,故$CF=sqrt{2}DE$;(3)设$DE=x$可得:$FE^{2}=AE^{2}=AM^{2}+ME^{2}=left(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}$,而$CF=sqrt{2}DE$,若$FE$$=FC$,有$left(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}=(sqrt{2}x)^{2}$,可解得$DE=2sqrt{6}-2sqrt{2}$.【解析】(1)证明:过点$E$作$MNperpAD$于$M$,交$BC$于点$N$,如图:<ImageHere>$because$四边形$ABCD$为正方形,$thereforeAD//BC,AD=DC,angleADB=45^{circ}$,$becauseMNperpAD$,$thereforeMNperpBC$,$therefore$四边形$NCDM$为矩形,$thereforeMN=CD$,$becauseangleADB=45^{circ},MNperpAD$,$thereforeMD=ME$,$thereforeAM=EN$,$becauseAEperpEF$,$thereforeangleAEM+angleFEN=90^{circ}$.$becauseangleAEM+angleMAE=90^{circ}$,$thereforeangleFEN=angleMAE$,$thereforetriangleAEMcongtriangleEFN(ASA)$$thereforeAE=EF$.(2)解:$CF=sqrt{2}DE$,理由如下:由(1)知$triangleAEMcongtriangleEFN,angleADB=45^{circ}$,$thereforeME=FN=MD$,$because$四边形$NCDM$为矩形,$thereforeCN=MD$,$thereforeCF=2MD$,$becauseDE=sqrt{2}MD$,$thereforeCF=sqrt{2}DE;$(3)解:设$DE=x$.由(1)得:$FE^{2}=AE^{2}=AM^{2}+ME^{2}=left(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}$,由(2)得$CF=sqrt{2}DE$,$thereforeCF=sqrt{2}x$,$becauseFE=FC$,$thereforeFE^{2}=FC^{2}$,$thereforeleft(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}=(sqrt{2}x)^{2}$,解方程得:$x_{1}=2sqrt{6}-2sqrt{2},x_{2}=-2sqrt{6}-2sqrt{2}$(舍去),$thereforeDE=2sqrt{6}-2sqrt{2}$.【点睛】本题考查正方形性质及应用,涉及全等三角形的判定与性质,勾股定理及应用等,解题的关键是作辅助线,构造三角形全等.
解析：(1)过点$E$作$MNperpAD$于$M$,交$BC$于点$N$,由四边形$ABCD$为正方形,$AEperpEF$,可证明$triangleAEMcongtriangleEFN(ASA)$，即可得$AE=EF$.(2)由$triangleAEMcongtriangleEFN,angleADB=45^{circ}$，可得$CF=2MD$,而$DE=sqrt{2}MD$,故$CF=sqrt{2}DE$;(3)设$DE=x$可得:$FE^{2}=AE^{2}=AM^{2}+ME^{2}=left(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}$,而$CF=sqrt{2}DE$,若$FE$$=FC$,有$left(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}=(sqrt{2}x)^{2}$,可解得$DE=2sqrt{6}-2sqrt{2}$.【解析】(1)证明:过点$E$作$MNperpAD$于$M$,交$BC$于点$N$,如图:$because$四边形$ABCD$为正方形,$thereforeAD//BC,AD=DC,angleADB=45^{circ}$,$becauseMNperpAD$,$thereforeMNperpBC$,$therefore$四边形$NCDM$为矩形,$thereforeMN=CD$,$becauseangleADB=45^{circ},MNperpAD$,$thereforeMD=ME$,$thereforeAM=EN$,$becauseAEperpEF$,$thereforeangleAEM+angleFEN=90^{circ}$.$becauseangleAEM+angleMAE=90^{circ}$,$thereforeangleFEN=angleMAE$,$thereforetriangleAEMcongtriangleEFN(ASA)$$thereforeAE=EF$.(2)解:$CF=sqrt{2}DE$,理由如下:由(1)知$triangleAEMcongtriangleEFN,angleADB=45^{circ}$,$thereforeME=FN=MD$,$because$四边形$NCDM$为矩形,$thereforeCN=MD$,$thereforeCF=2MD$,$becauseDE=sqrt{2}MD$,$thereforeCF=sqrt{2}DE;$(3)解:设$DE=x$.由(1)得:$FE^{2}=AE^{2}=AM^{2}+ME^{2}=left(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}$,由(2)得$CF=sqrt{2}DE$,$thereforeCF=sqrt{2}x$,$becauseFE=FC$,$thereforeFE^{2}=FC^{2}$,$thereforeleft(4-frac{sqrt{2}}{2}xright)^{2}+left(frac{sqrt{2}}{2}xright)^{2}=(sqrt{2}x)^{2}$,解方程得:$x_{1}=2sqrt{6}-2sqrt{2},x_{2}=-2sqrt{6}-2sqrt{2}$(舍去),$thereforeDE=2sqrt{6}-2sqrt{2}$.【点睛】本题考查正方形性质及应用,涉及全等三角形的判定与性质,勾股定理及应用等,解题的关键是作辅助线,构造三角形全等.