1、证明：连接OC∵AC平分∠BAD∴∠BAC＝∠DAC∵OA＝OC∴∠ACO＝∠BAC∴∠ACO＝∠DAC∵CE⊥AD∴∠ACE+∠DAC＝90∴∠OCE＝∠ACE+∠ACO＝∠ACE+∠DAC＝90∴CE是圆O的切线2、∵CE是圆O的切线∴∠DAC＝∠DCE∵∠AEC＝∠CED＝90∴△ACE∽△CDE∴CE/DE＝AE/CE∵AE＝AD+DE＝3∴CE/1＝3/CE∴CE²＝3∴AC＝√（AE²+CE²）＝√（9+3）＝2√
1、证明：连接OC∵AC平分∠BAD∴∠BAC＝∠DAC∵OA＝OC∴∠ACO＝∠BAC∴∠ACO＝∠DAC∵CE⊥AD∴∠ACE+∠DAC＝90∴∠OCE＝∠ACE+∠ACO＝∠ACE+∠DAC＝90∴CE是圆O的切线2、∵CE是圆O的切线∴∠DAC＝∠DCE∵∠AEC＝∠CED＝90∴△ACE∽△CDE∴CE/DE＝AE/CE∵AE＝AD+DE＝3∴CE/1＝3/CE∴CE²＝3∴AC＝√（AE²+CE²）＝√（9+3）＝2√
