如图,三角形ABC中,AB=BC,BE⊥AC于点E,AD⊥BC于点D,角BAD=45°,AD与BE交于点F,连接CF

∵AD⊥BC,∠BAD=45°,∴⊿ADB是等腰直角三角形,AD=BD；
∵AB=BC,BE⊥AC,∴AE=EC,AC=2AE,
∵Rt⊿EBC与Rt⊿DAC有公用锐角∠C,∴∠EBC=∠DAC,
可证Rt⊿DBF≌Rt⊿DAC,得BF=AC=2AE.
BE⊥AC AB=BC ∠ABE=∠CBE 即∠ABF=∠CBF
AB=CB ∠ABF=∠CBF BF=BF △ABF≌△CBF
AF=CF
∠BAD=45 ∠ADB=90 ∠ABD=45
∠BAD=(180-45)/2=67.5 ∠DAC=67.5-45=22.5
∠DBF=45/2=22.5
∠ABD=∠BAD BD=AD ∠DBF=∠DAC=22.5
RT△BDF≌RT△ADC DF=DC=√2
∠FDC=90 AF=CF=√2CD=2
AD=AF+DF=2+√2