Question

如图,△ABC中,∠BAC=90°,角平分线BE与高AD相交于点F,求证：AE=AF

Answer 1

过E点做EM垂直BC于M,

∠ABE=∠CBE,EM⊥BC,EA⊥AB,AE=ME,△ABE≌△MBE,∠AEB=∠MEB

AD⊥BC,EM⊥BC,AD || EM,∠AFE=∠MEB,∠AFE=∠AEB,AE=AF.