解:(1)AB=AC说理如下: ∵EC平分∠AED,DB平分∠ADE, ∴∠AEC=∠AED,∠ADB=∠ADE. ∵∠AED=∠ADE, ∴∠AEC=∠ADB.在△AEC和△ADB中,∠AEC=∠ADB,AE=AD,∠A=∠A, ∴△AEC≌△ADB(ASA) ∴AB=AC; (2)BE=CD,BE⊥CD ∵∠EAD=∠BAC, ∴∠EAD+∠BAD=∠BAC+∠BAD, ∴∠EAB=∠DAC,在△AEB和△ADC中,AB=AC∠EAB=∠DACAE=AD ∴△AEB≌△ADC(SAS), ∴∠AEB=∠ADC, ∵∠AEB+∠DEB+∠ADE=90°, ∴∠ADC+∠DEB+∠ADE=90°①, ∵∠ADC+∠DEB+∠ADE+∠DOE=180°②, ②﹣①得,∠DOE=90°, ∴BE⊥CD. |