如图,△ABC是边长为4CM的三角形,P是△ABC内的任意一点,过点P作EF‖AB分别交AC,BC于点E,F,作GH‖B

应该是边长为4CM的“正”三角形吧

∵EF‖AB,GH‖BC,MN‖AC
∴四边形AMPE,BGPF,CNPH都是平行四边形
AM=EP,AE=MP,BG=FP,BF=GP,CN=HP,CH=NP
且△ABC∽△MGP∽△EPH∽△PFN
设MP/AC=GP/BC=MG/AB=k1
EH/AC=PH/BC=EP/AB=k2
PN/AC=FN/BC=PF/AB=k3
∴k1+k2+k3
=MP/AC+EH/AC+PN/AC
=(MP+EH+PN)/AC
=(AE+EH+HC)/AC
=AC/AC
=1
则MP/AC+GP/BC+PH/BC+EP/AB+PN/AC+PF/AB
=(MP+PN)/AC+(GP+PH)/BC+(EP+PF)/AB
=MN/AC+GH/BC+EF/AB
=2(k1+k2+k3)
=2
∵△ABC是边长为4CM的正三角形
AC=BC=AB=4
∴MN/AC+GH/BC+EF/AB=(MN+GH+EF)/AC=2
MN+GH+EF=2AC=2×4=8(cm)