ab=ac.ab=ae角bac=角dae点c在de上求证.角ab全等角ace

∵DE平行于AB交AC与E∴△CED∽△CAB,∠EDA=∠DAB∴CE：AC=DE：AB∵AD为三角形ABC的角平分线∴∠EAD=∠DAB=∠EDA∴AE=DE∴CE：AC=DE：AB=AE：AB∴

∵∠BAC=2∠B(∠ABC),AE平分∠BAC,∴∠CAE=∠BAE=∠ABC∴∠CEA=∠BAC,∵∠ACE=∠ACB,∴△ACE∽△ABC,∵AB=2AC,∴AE=2CE

证明：（话说那个“AE平分角BAC”没用）在三角形ABC中,∠BAC=2倍的∠B,AB=2AC,取AB的中点D,连接CD,则有AD=AC=BD,所以∠ACD=∠ADC,∠DCB=∠B,∠C=∠ACD+

角B=角BAE,推出AE=BE,取AB的中点为H,连接EH,根据AE=BE,中点为H,所以EH垂直于AB,AH=0.5AB=AC,角BAE=角CAE,所以三角形AEH全等于三角形ACE,后面不用了吧

作角a的平分线AD,交BC于D,再取AB的中点E,连接DEAC=0.5AB=AE角EAD=角CAD,所以△EAD全等△CAD所以角c=角AED,角EAD=角CAD=0.5角BAC=角B,所以三角ABD

证明：∵∠DAE=∠BAC,∴∠DAE-∠CAD=∠BAC-∠CAD,即∠EAC=∠DAB,∵AE=AD,AC=AB,∴ΔAEC≌ΔADB,∴CE=BD.(注：不是CE=BC).

解答证明：∵∠BAC=∠DAE,∴∠BAC+∠CAD=∠DAE+∠CAD,即∠BAD=∠EAC,在△ABD和△ACE中AB=AC∠BAD=∠EACAE=AD,∴△ABD≌△ACE．所以∠ADB=∠AE

有图吗,有图就好做

证明：设AB＝AC＝3X,过点E作EF⊥BC于F∵∠BAC＝90,AB＝AC＝3X∴∠ABC＝∠C＝45,BC＝3√2X∵AE＝1/3AC∴AE＝X∴CE＝AC-AE＝2X∵EF⊥BC∴CF＝EF＝C

因为AB=AC,AD=AE,BD=CE所以△ABD≌△ACE所以∠BAD=∠CAE所以∠BAD+DAC=∠DAC+∠CAE所以角BAC=角DAE

图呢.=.==.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=

因为AB=AC,BD=CE,AD=AE所以△ABD≌△ACE所以∠BAD=∠CAE又∠BAC=∠BAD+∠CAD,∠DAE=∠CAE+∠CAD所以∠BAD=∠DAE

再答：

AD平分∠BAC,角1=角2DE‖AC,角2=角ADEAE=DEBE/AB=DE/ACBE/AB=AE/AC(AB-AE)/AB=AE/AC1-AE/AB=AE/ACAE/AB+AE/AC=1

∵AB=AC∴∠B=∠ACB∵AB‖DE∴∠B=∠EDC∴∠EDC=∠ACB∵∠FAE=∠EAC,AB‖DE∴∠FAE=∠AED,∠EAC=∠AED∵ABC为等腰三角形∴∠BAD=∠CAD,BD=DC

证明：因为在三角形ABC中AB=AC并且AD是高所以AD⊥BC∠ADC=90°∠DAC=1/2∠BAC又因为AE平分∠MAC所以∠FAC=1/2∠CAM所以∠DAF=1/2×180°=90°因为DF‖

因为AD=AE,AB=AC,∠BAD=∠CAE所以△ADB≌△AEC所以∠ADB=∠AEC,BD=CE因为BD=CE,DE=BC所以四边形BCED是平行四边形所以BD=CE所以∠BDE+∠DEC=18

1.因为∠BAC=∠DAE所以∠BAC+∠DAC=∠DAE+∠DAC即∠BAD=∠CAE因为AB=AC,AD=AE所以△ABD≌△ACE（SAS）2.AC与BD相交于O点,在△BOA和△COF中因为△

对了就是9再问：我想要过程再答：AB=AC∠BAC=120°∴∠B=∠C=30°∠BAE=90°∴BE=2AE=6∴∠EAC=30°=∠C∴EA=EC=3∴BC=BE+EC=9

因为AB=2AC,所以∠B=30度因为∠BAC=2∠B,AE平分∠CAB所以∠BAE=∠CAE=∠B=30度,BE=AE所以CE=二分之一AE=二分之一BE即AE=2CE