分类：教育/科学-学习帮助
问题：几何 
回答：(1)过D作DG垂直BC于G,连FGAE⊥BC,∠FBC=30==>∠AFO=∠BFE=60AD∥BC,DG⊥BC,AE⊥BC==>AD=EGDG⊥BC,BF=FD==>BF=FG==>BE= :3==>BE=EG=GCAD∥BC,BF=FD==>AF=FE==>FG//AC==>∠FAO=∠EFGBF=FG,FE⊥BG,∠FBC=30==>∠EFC=60==>∠FAO=60==>△AOF为等边三角形(2)AF=OF=FE=1/2*BF,BF=2OFBF+OF=3OF=k-2BF*OF=2OF^2=k3OF=2OF^2-2,2OF^2-3OF-2=0OF=2k=3OF+2=8AF=FE=OF=2BF=2OF=4BE=2√3AD=BE=2√3BC=3BE=6√3S=1/2(AD+BC)*AE=1/2*8√3*4=16√3