设D为AB中点,则
OA+AB=OB,AB=OA-OB,同理CA=OC-OA,CD=CA+AD,
OG=OC+CG= OC+2/3CD= OC+2/3(CA+AD)
=OC+2/3(CA+1/2AB)= OC+2/3CA+1/3AB
=OC+2/3(OC-OA)+1/3(OA-OB)
=5/3OC-1/3OA-1/3OB
OG=5/3OC-1/3OA-1/3OB
同理
OG=5/3OA-1/3OB-1/3OC
OG=5/3OB-1/3OC-1/3OA
上面三式相加得
3OG=OA+OB+OC
OG=(OA+OB+OC)/3