题目（九年级 · 度量几何学）：(本题10分)(2022-浙江温州$cdot$一模)如图,在$RttriangleABC$中,$angleABC=90^{circ},D$是$BC$上的一点,且$angleBAD=angleACB,DEperpAC$于点$F$,交$BC$的平行线$AE$于点$E$.(1)求证:$AD=DE$.(2)若$BD=frac{sqrt{15}}{3},CD=sqrt{15}$.(1)求$AC$的长.(2)过点$E$作$EGperpAD$于点$G$,在射线$AC$上取一点$M$与$triangleAEG$某一边的两端点,构成以$M$为顶点的角等于$angleACB$,求所有满足条件的$AM$的长.
选项：
答案：(1)见解析,(2)(1)$frac{10sqrt{3}}{3}$,(2)$frac{8}{3}sqrt{3}$或$frac{4}{3}sqrt{3}$或$frac{10sqrt{3}}{3}$
解析：【分析】(1)因为$AE//BC$,所以有$angleBAE+angleABC=180^{circ},angleACB=angleCAE$,又因为$angleBAD=angleACB$,得用直角三角形两锐角互余得出$angleDAE=angleE$,即可得出结论;（2）(1)证$triangleABDsimtriangleCBA$,得$frac{AB}{BC}=frac{BD}{AB}$,即可求出$AB^{2}$值,再由勾股定理求$AC$长即可;(2)分情况讨论：I）当$angleAME=angleACB$时,II）当$angleAMG=angleACB$时,III）当$angleEMG=angleACB$时,分别求解即可.(1)证明:$becauseAE//BC$,$thereforeangleBAE+angleABC=180^{circ},angleACB=angleCAE$,$becauseABC=90^{circ}$,$thereforeangleBAE=90^{circ}$，即$angleBAD+angleDAE=90^{circ}$,$becauseangleBAD=angleACB$$thereforeangleBAD=angleCAE$,$becauseDEperpAC$$thereforeangleAFE=90^{circ}$,$thereforeangleCAE+angleE=90^{circ}$,$thereforeangleDAE=angleE$,$thereforeAD=DE$(2)解:(1)$becauseangleBAD=angleACB,angleABD=angleCBA$,$thereforetriangleABDsimtriangleCBA$,$thereforefrac{AB}{BC}=frac{BD}{AB}$$thereforeAB^{2}=BDcdotBC=frac{sqrt{15}}{3}left(frac{sqrt{15}}{3}+sqrt{15}right)=frac{20}{3}$,$thereforeAC^{2}=AB^{2}+BC^{2}=frac{20}{3}+left(frac{sqrt{15}}{3}+sqrt{15}right)^{2}=frac{100}{9}$,$thereforeAC=frac{10sqrt{3}}{3}$;(2)分三种情况:$I)$当$angleAME=angleACB$时,如图,$becauseangleACB=angleEAC$$thereforeangleEAC=angleAME$,$thereforeAE=ME$,$becauseEFperpAC$$thereforeAM=2AF$$becauseS_{triangleACD}=frac{1}{2}ACcdotDF=frac{1}{2}CDcdotAB$,$thereforefrac{1}{2}timesfrac{10sqrt{3}}{3}cdotDF=frac{1}{2}timessqrt{15}timessqrt{frac{20}{3}}$,$thereforeDF=sqrt{3}$,在$RttriangleABD$中,$AD=sqrt{AB^{2}+BD^{2}}=sqrt{frac{20}{3}+left(frac{sqrt{15}}{3}right)^{2}}=frac{5}{3}sqrt{3}$,在$RttriangleADF$中,$AF=sqrt{AD^{2}-DF^{2}}=sqrt{left(frac{5sqrt{3}}{3}right)^{2}-(sqrt{3})^{2}}=frac{4}{3}sqrt{3}$,$thereforeAM=2AF=frac{8}{3}sqrt{3}$;II)当$angleAMG=angleACB$时,如图,$thereforeGM//BC$,$thereforetriangleAGMsimtriangleADC$,$thereforefrac{AM}{AC}=frac{AG}{AD}$,在$triangleADF$和$triangleEDG$中,$left{begin{array}{l}angleAFD=angleEGDangleADF=angleEDG,AD=EDend{array}right.$$thereforetriangleADFcongtriangleEDG({AAS})$,$thereforeDG=DF=sqrt{3}$,$thereforeAG=frac{2}{3}sqrt{3}$,$thereforefrac{AM}{frac{10sqrt{3}}{3}}=frac{frac{2}{3}sqrt{3}}{frac{5}{3}sqrt{3}}$,$thereforeAM=frac{4}{3}sqrt{3}$.III）当$angleEMG=angleACB$时,如图所示,过$E$作$EHperpBC$于$H$,延长$AD$交$EH$于$Q$,$becauseAE//BC,ABperpBC,EHperpBC$$thereforeAB=EH$,又$AD=DE$$thereforetriangleABDcongtriangleEDH$$thereforeBD=DH=frac{sqrt{15}}{3}$,$thereforeBH=frac{2sqrt{15}}{3},CH=CD-DH=frac{2sqrt{15}}{3}$又$AB=frac{2sqrt{15}}{3}=EH$,$thereforeEH=BH=CH$在$triangleABD$和$triangleQHD$中,$left{begin{array}{l}angleADB=angleQDHBD=HDangleABD=angleQHD=90^{circ}end{array}right.$$thereforetriangleABDcongtriangleQHD({ASA})$$thereforeAB=HQ=CH$,故$H$为$EQ$的中点,又$EGperpAD$即$GH$为直角三角形$EGQ$斜边中线$thereforeGH=EH=HQ$$thereforeGH=EH=HQ=CH$即$E、G、Q、C$四点共圆，如下图:$thereforeangleECG=angleEQG($同弧所对的圆周角相等$)$由$triangleABDcongtriangleQHD$知,$angleEQG=angleBAD=angleACB$$thereforeangleECG=angleACB$,当$angleEMG=angleACB$,点$M$在射线$AC$上时,即此时$M$与$C$重合,则$AM=AC=frac{10sqrt{3}}{3}$.【点睛】本题考查平行线的性质,相似三角形的判定与性质,勾股定理,等腰三角形的性质,全等本角形的判定与性质,注意（2）(2)要分类讨论，以免漏解.