1、设△ABC,BC边上中线为AD=3cm,AC边上中线BE=4cm,AB边上中线CF=5cm,重心G,延长GD至M,使DM=GD,连结CM,BM,四边形BMCG是平行四边形,根据重心性质,GD=AD/3=1cm,则GM=2cm,CM=BG=8/3cm,CG=10/3cm,根据勾股定理逆定理,△GMC是直角三角形,CD^2=DM^2+CM^2,,CD=√(1+64/9)=√73/3cm,BC=2CD=2√73/3cm,同理延长GE至N,使EN=GE,连CN,AN,CE=√(2^2+16/9)= 2√13/3(cm),AC=2CE=4√13/3(cm),同理延长GF至P,使FP=FG,则<PBG=90度,BF=PG/2=5 |