三角形ABC中,AB =6,AC=9,D在AC上,且AD=4,E ,F分别是BD,BC中点,AE=3,求AF的长

连接FE,FE延长线交AB于G,
E ,F分别是BD,BC中点,
BE:BD=EF:DC=1:2,
EF=0.5*DC=0.5*(AC-AD)=0.5*(9-4)=2.5;
BE:BD=BG:BA=1:2,
BG=0.5*BA=0.5*6=3,AG=BG=3=AE,[已知AE=3]
BE:BD=GE:AD=1:2,
GE=0.5*AD=0.5*4=2;
AG=AE,三角形AGE为等腰三角形;作AH垂直于GE,垂足H,GH=EH=GE/2=2/2=1,
AH²=AE²-EH²=3²-1²=8;
AF²=AH²+HF²=8+(HE+EF)²=8+(1+5/2)²=8+49/4=81/4,
AF=√(81/4)=9/2