如图,点P为△AEF外一点,PA平分∠EAF,PD⊥EF于D,且DE=DF,PB⊥AE于B. 求证:AF-AB=BE.
问题描述:
如图,点P为△AEF外一点,PA平分∠EAF,PD⊥EF于D,且DE=DF,PB⊥AE于B.
求证:AF-AB=BE.
答
证明:如图,过点P作PM⊥AF于M,连接PE、PF,∵PA平分∠EAF,PB⊥AE,∴PB=PM,AM=AB,∵PD⊥EF,DE=DF,∴PD垂直平分EF,∴PE=PF,在Rt△PBE和Rt△PMF,PE=PFPB=PM,∴Rt△PBE≌Rt△PMF(HL),∴MF=BE,∵AF-AM=MF...