如图1,在Rt△ABC中,∠ACB=90°,D是AB上的一点,且∠ACD=∠B.（1）求证：CD⊥AB

①证明：∵∠ACB=90°∴∠A+∠B=90°∵∠ACD=∠B∴∠A+∠ACD=90°则∠ADC=90°即CD⊥AB②证明∵∠ACE=∠ADF=90°∴∠CAE+∠AEC=∠DAF+∠AFD=90°∵AE平分∠BAC∴∠CAE=∠DAF∴∠AEC=∠AFD∵∠CFE=∠AFD（对顶角相等）∴∠AEC=∠CFE③∵∠ACB=∠ADC=90° ∠BAC=∠CAD∴△ACB∽△ADC（AA）∴AC/AD=AB/AC∴AC²=AD×AB=4AD²AC=2AD则∠B=∠ACD=30°∵BC=3CE∴BE=2CE∵AB=2AC即AB/AC=BE/CE∴AE平分∠BAC∴AC/AD=CF/DF =2即DF=1/3CD∵S△ABC=½AB×CD=48 AD=1/4AB ∴S△ADC=½AD×CD=12 S△ADF=S△ADC/3=4 S△ACF=S△ADC -S △ADF=8∵CE=1/3BC∴S△ACE=S△ABC/3=16S△CEF=S△ACE-S△ACF=16-8=8