在直角三角形ABC中,AC=BC,AD是角CAB的角平分线,过点B作BE垂直AD,交AD的延长线于点E,

延长AC、BE交于F∵Rt△ABC中AC=BC∴∠CAB=∠CBA=45°∵AD平分∠CAB∴∠CAD=22.5°∵AE⊥BE即AE⊥BF且AD平分∠CAB即AE平分∠FAB∴AE为BF中垂线∴AB=AF(中垂线上的点到线段两端的距离相等)∴∠F=∠ABF=(80°-45°)/2=67.5°∴∠FBC=∠ABF-∠CBA=67.5°-45°=22.5°=∠CAD在△ADC和△CBF中∠CAD=∠FBCAC=CB ∠ACB=∠FCB=90°∴△ADC全等△CBF∴BF=AD∵AE为BF中垂线∴BE=BF=1/2BF∵BF=AD∴BE=1/2AD啊~我从来没有打过那么复杂的题~~~