證明：連接AP，BP，CP，
∵PE⊥AB，PF⊥AC，PD⊥BC，AH⊥BC于H，
∴S_△ABC=

1

2

BC?AH，S_△APB=

1

2

AB?PE，S_△APC=

1

2

AC?PF，S_△BPC=

1

2

BC?PD
∵S_△ABC=S_△APB+S_△APC+S_△BPC
∴

1

2

BC?AH=

1

2

AB?PE+

1

2

AC?PF+

1

2

BC?PD，且AB=BC=AC，
即PE+PF+PD=AH．