如图,在Rt△ABC中,∠BAC=90°,AD⊥BC,D为垂足,∠ACD的平分线交AD.AB于点E,F则AE=AF,请你

证明：AD⊥BC,AB⊥AC.∠B+∠BAD=90°∠DAC+∠BAD=90°∠B=∠DAC
∠AEF=∠DAC+∠ACE,∠AFE=∠B+∠FCB ∠ACE=1/2∠ACB=∠FCB
∴∠AEF=∠AFE.AF=AE