设OE=x,OF=y,BC=a,根据重心的性质,BO=2OE,CO=2OF,BE⊥CF,△BOF,△COE,△BOC均是RT△,BO^2+OF^2=BF^2,CO^2+OE^2=CE^2,BO^2+CO^2=BC^2,x^2+(2y)^2=(b/2)^2,x^2+4y^2=b^2/4,........(1),4x^2+y^2=c^2/4,.........(2),4x^2+4y^2=a^2,...........(3),(1)+(2)得:,5(x^2+y^2)=(b^2+c^2)/4,x^2+y^2=(b^2+c^2)/20,a^2=4(x^2+y^2)=(b^2+c^2)/5,∴a=√5(b^2+c^2)/5.
|