仅提供思路：
连CH,并延长,交AB与D
=>
CH⊥AB
PA,PB,PC两两垂直
PC⊥PA,PC⊥PB
=>
PC⊥PAB
=>
PC⊥AB
=>
AB⊥PCH
=>
AB⊥PH
同理
BC⊥PH
AC⊥PH
=>
PH⊥ABC