应该是AD=2BE
延长AC和BE,交于F
∵BE⊥AD,那么∠AEB=∠AEF=90°
AD平分∠BAC,那么∠BAE=∠FAE(∠CAE)
AE=AE
∴△ABE≌△AFE(ASA)
∴EF=BE=1/2BF,即BF=2BE
∵∠ACB=∠AEF=90°
∴∠CBF=90°-∠F
∠CAD=∠FAE=90°-∠F
那么∠CBF=∠CAD
∵CA=CB,∠ACD=∠BCF=90°
∴△ACD≌△BCF(ASA)
∴AD=BF=2BE
应该是AD=2BE
延长AC和BE,交于F
∵BE⊥AD,那么∠AEB=∠AEF=90°
AD平分∠BAC,那么∠BAE=∠FAE(∠CAE)
AE=AE
∴△ABE≌△AFE(ASA)
∴EF=BE=1/2BF,即BF=2BE
∵∠ACB=∠AEF=90°
∴∠CBF=90°-∠F
∠CAD=∠FAE=90°-∠F
那么∠CBF=∠CAD
∵CA=CB,∠ACD=∠BCF=90°
∴△ACD≌△BCF(ASA)
∴AD=BF=2BE