AO·BC是個定值,與BC都沒有關系
令外接圓半徑為：R
即：|OA|=|OB|=|OC|=R
AO·BC=AO·(OC-OB)
=OA·OB-OA·OC
=R^2cos-R^2cos
=R^2(2R^2-|AB|^2)/(2R^2)-R^2(2R^2-|AC|^2)/(2R^2)
=R^2-|AB|^2/2-R^2+|AC|^2/2
=|AC|^2/2-|AB|^2/2
=9/2-2=5/2
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或者：取AB邊中點D,AC邊中點E
則：OD⊥AB,OE⊥AC
故：cos(∠OAB)=|AD|/|AO|=1/R
cos(∠OAC)=|AE|/|AO|=3/(2R)
AO·BC=AO·(AC-AB)
=AO·AC-AO·AB
=R*|AC|cos(∠OAC)-R*|AB|(∠OAB)
=3*R*3/(2R)-R*2/R
=9/2-2
=5/2