已知：△ABC中,AB=AC,BD是AC的中线,△ABD周长为21,△BDC周长为12,求AB,BC的长度

设:AB=2X,则:AC=2X,AD=DC=X △ABD周长-△BDC周长=AB-BC+AD-DC=AB-BC=2X-BC=21-12=9 BC=2X-9 作AE垂直BC于E,交BD于F,则:DE=EC=DC/2 则:F为三角形ABC的重心(三条中线的交点) 所以:EF=AE/3,BF=(2/3)BD 而:BF^2=EF^2+BE^2 (4/9)BD^2=(1/4)BC^2+(1/9)AE^2=(1/4)BC^2+(1/9)(AB^2-BE^2) =(1/4)BC^2+(1/9)(AB^2-(1/4)BC^2) =(2/9)BC^2+(1/9)AB^2 =(2/9)(2X-9)^2+(1/9)(2X)^2 BD^2 =3X^2-18X+(81/2) 而:AB+AD+BD=21 BD=21-3X 所以:3X^2-18X+(81/2)=(21-3X)^2 6X^2-108X+(801/2)=0 4X^2-72X+267=0 X=9-((根号57)/2) (另一根9+((根号57)/2)应舍弃,因X