已知等腰△ABC中,AB+BC,P在底边AC上的任一点,PE⊥AB于点E,PF⊥BC于点F,CD⊥AB于点D.求证：CD=PE+PF

这里应该是AB=BC把.连接BP,有了俩个三角形!S△ABP=1/2(AB×PE),S△BCP=1/2(BC×pF),S△ABC=1/2(AB×CD)∵AB=BC,∴1/2(AB×PE)+1/2(BC×PF)=1/2(AB×CD),∴1/2×AB×(PE+PF)=1/2×AB×CD,(把bc换成ab,∴CD=PE+PF 采纳...