1.(1)连结AO,EO,延长OF至点F'使OF=FF',又AF=FC,有四边形AOCF'是平行四边形,故AO//CF',AF'//CD,CO=AF',AO=CF'所以三角形BOD相似于BF'A,则BO/BF'=BD/BA=1/2,DO/AF'=BD/BA=1/2,即CO=AF'=2DO,同理CO=2DO又BE/BC=1/2=BO/BF',所以EO//CF'//AO,EO=CF'/2=AO/2,则有A,O,E三点共线,AO=2OE(2)向量AO=AE*2/3=(AB/2+AC/2)*2/3=(AB+AC)/32.(1)向量AB=SB-SA=(-2,7),故AB=根号53.有SA=5,cosSAB=AB点乘SA/(AB*SA)=(-4...
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