(1)AB=AC
理由如下:∵EC平分∠AED,DB平分∠ADE,
∴∠AEC= 1/2∠AED,∠ADB= 1/2∠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=∠DAC
AE=AD
∴△AEB≌△ADC(SAS),
∴∠AEB=∠ADC,
∵∠AEB+∠DEB+∠ADE=90°,
∴∠ADC+∠DEB+∠ADE=90°,
∵∠ADC+∠DEB+∠ADE+∠DOE=180°,
∴∠DOE=90°,
∴BE⊥CD.