证明：连接AP，BP，CP，∵PE⊥AB，PF⊥AC，PD⊥BC，AH⊥BC于H，∴S△ABC=12BC•AH，S△APB=12AB•PE，S△APC=12AC•PF，S△BPC=12BC•PD∵S△ABC=S△APB+S△APC+S△BPC∴12BC•AH=12AB•PE+12AC•PF+12BC•PD，且AB=...