证明 设AF/FB=BD/DC=CE/EA=m/n，则S(AEF)/S(ABC)=AE*AF/AB*AC=mn/(m+n)^2;同理得:S(BDF)=S(CED)=S(AEF)=[mn/(m+n)^2]*S(ABC). (1)令G为△ABC的重心，连结GA,GB,GC;GD,GE,GF.则S(AFG)/S(ABG)=AF/AB=m/(m+n),S(AFG)=[m/(m+n)]*S(ABG)=(1/3)*[m/(m+n)]*S(ABC)同理得:S(BDG)=S(CEG)=S(AFG)=(1/3)*[m/(m+n)]*S(ABC) (2)S(BFG)=S(CDG)=S(AEG)=(1/3)*[n/(m+n)]*S(ABC) (3)(2)+(3)得:S(BDGF)=S(CEGD)=S(AFGE)=S(ABC)/3 (4).(4)-(1)得:S(FGD)=S(DGE)=S(EGF)=(1/3)*[(m^2-mn+n^2)/(m+n)^2]*S(ABC).根据三角形重心定理，即知G为△DEF的重心.