题目（八年级 · 度量几何学）：如图1,四边形$ABCD$是正方形,点$E$是边$BC$的中点,$angleAEF=90^{circ}$,且$EF$交正方形外角的平分线$CF$于点$F$,过点$F$作$FGperpBC$于点$G$,连接$AC$.易证:$AC=sqrt{2}(EC+FG)$.(提示:取$AB$的中点$M$,连接$EM)$(1)当点$E$是$BC$边上任意一点时,如图2;当点$E$在$BC$延长线上时,如图3.请直接写出$AC$,$EC,FG$的数量关系,并对图2进行证明;(2)已知正方形$ABCD$的面积是27,连接$AF$,当$triangleABE$中有一个内角为$30^{circ}$时,则$AF$的长为$6sqrt{2}$或$6sqrt{6}ldots$.图1图2图3
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答案：(1)如图2中,结论:$AC=sqrt{2}(FG+EC)$.在$AB$上截取$BM=BE$,连接$EM$,证明$triangleAEMcongtriangleEFC(ASA)$,可得结论.如图3中,结论:$AC=sqrt{2}(FG-EC)$.（2）分两种情形：如图1中,当$angleBAE=30^{circ}$时,如图3中,当$angleAEB=30^{circ}$时,利用等腰直角三角形的性质求解即可.【解析】解：(1)如图2中,结论:$AC=sqrt{2}(FG+EC)$.理由:在$AB$上截取$BM=BE$,连接$EM$,$because$四边形$ABCD$是正方形,$thereforeangleB=angleBCD=90^{circ},AB=BC$$thereforeangleDCG=90^{circ},angleEAM+angleAEB=90^{circ}$,$becauseBM=BE$,$thereforeAB-BM=BC-BE,angleBME=angleBEM=45^{circ}$,$thereforeAM=EC,angleAME=135^{circ}$,$becauseCF$平分$angleDCG$,$thereforeangleFCG=45^{circ}$,$thereforeangleECF=135^{circ}$,$thereforeangleAME=angleECF$,$becauseangleAEF=90^{circ}$$thereforeangleFEC+angleAEB=90^{circ}$,$thereforeangleEAM=angleFEC$,$therefore$在$triangleAEM$和$triangleEFC$中,<ImageHere>$thereforetriangleAEMcongtriangleEFC(ASA)$,$thereforeEM=CF$,$becauseEM=sqrt{2}BE,quadCF=sqrt{2}FG$,$thereforeBE=FG$,$becauseAC=sqrt{2}BC=sqrt{2}(BE+EC)$,$thereforeAC=sqrt{2}(FG+EC)$.如图3中,结论:$AC=sqrt{2}(FG-EC)$.(2)如图1中,当$angleBAE=30^{circ}$时,$because$正方形的面积为27,$thereforeAB=3sqrt{3},angleB=90^{circ}$,$thereforeBE=ABcdottan30^{circ}=3sqrt{3}timesfrac{sqrt{3}}{3}=3$,$thereforeAE=2BE=6$,$becausetriangleAEMcongtriangleEFC$$thereforeAE=EF=6$,$thereforeAF=6sqrt{2}$,如图3中,当$angleAEB=30^{circ}$时,同法可得$AE=EF=2AB=6sqrt{3}$,$thereforeAF=sqrt{2}AE=6sqrt{6}$,综上所述,$AF$的长为$6sqrt{2}$或$6sqrt{6}$.<ImageHere>图2【点睛】本题考查正方形的性质,全等三角形的判定和性质,等腰直角三角形的性质等知识,解题的关键是学会添加常用辅助线,构造全等三角形解决问题,学会用分类讨论的思想思考问题.
解析：(1)如图2中,结论:$AC=sqrt{2}(FG+EC)$.在$AB$上截取$BM=BE$,连接$EM$,证明$triangleAEMcongtriangleEFC(ASA)$,可得结论.如图3中,结论:$AC=sqrt{2}(FG-EC)$.（2）分两种情形：如图1中,当$angleBAE=30^{circ}$时,如图3中,当$angleAEB=30^{circ}$时,利用等腰直角三角形的性质求解即可.【解析】解：(1)如图2中,结论:$AC=sqrt{2}(FG+EC)$.理由:在$AB$上截取$BM=BE$,连接$EM$,$because$四边形$ABCD$是正方形,$thereforeangleB=angleBCD=90^{circ},AB=BC$$thereforeangleDCG=90^{circ},angleEAM+angleAEB=90^{circ}$,$becauseBM=BE$,$thereforeAB-BM=BC-BE,angleBME=angleBEM=45^{circ}$,$thereforeAM=EC,angleAME=135^{circ}$,$becauseCF$平分$angleDCG$,$thereforeangleFCG=45^{circ}$,$thereforeangleECF=135^{circ}$,$thereforeangleAME=angleECF$,$becauseangleAEF=90^{circ}$$thereforeangleFEC+angleAEB=90^{circ}$,$thereforeangleEAM=angleFEC$,$therefore$在$triangleAEM$和$triangleEFC$中,$thereforetriangleAEMcongtriangleEFC(ASA)$,$thereforeEM=CF$,$becauseEM=sqrt{2}BE,quadCF=sqrt{2}FG$,$thereforeBE=FG$,$becauseAC=sqrt{2}BC=sqrt{2}(BE+EC)$,$thereforeAC=sqrt{2}(FG+EC)$.如图3中,结论:$AC=sqrt{2}(FG-EC)$.(2)如图1中,当$angleBAE=30^{circ}$时,$because$正方形的面积为27,$thereforeAB=3sqrt{3},angleB=90^{circ}$,$thereforeBE=ABcdottan30^{circ}=3sqrt{3}timesfrac{sqrt{3}}{3}=3$,$thereforeAE=2BE=6$,$becausetriangleAEMcongtriangleEFC$$thereforeAE=EF=6$,$thereforeAF=6sqrt{2}$,如图3中,当$angleAEB=30^{circ}$时,同法可得$AE=EF=2AB=6sqrt{3}$,$thereforeAF=sqrt{2}AE=6sqrt{6}$,综上所述,$AF$的长为$6sqrt{2}$或$6sqrt{6}$.图2【点睛】本题考查正方形的性质,全等三角形的判定和性质,等腰直角三角形的性质等知识,解题的关键是学会添加常用辅助线,构造全等三角形解决问题,学会用分类讨论的思想思考问题.