证明:∠BAC=180-(∠ABC+∠C)=180-4∠C∠1=∠BAC/2=90-2∠C∠ABE=90-∠1=2∠C延长BE交AC于或慧衫F因为,∠1 =∠2,BE⊥AE所以,△ABF是等腰三碧竖角形衫腔AB=AF,BF=2BE∠FBC=∠ABC-∠ABE=3∠C-2∠C=∠CBF=CFAC-AB=AC-AF=CF=BF=2BE