∵AD=xAB+yAC,又AD=AB+BD,
∴AB+BD=xAB+yAC,所以BD=(x-1)AB+yAC,
又∵AC⊥BD,∴BD·AB=(x-1)AB²,
设AB=1,则DE=BC=√2,
又∵∠BED=60º,∴BD=√6/2,
显然BD与AB夹角45°,∴BD·AB=(x-1)AB²,√6/2×1×cos45º=(x-1)×12,x=√3/2+1,
同理,BD=(x-1)AB+yAC,两边同时乘以AC,由数量积得y=√3/2.
∵AD=xAB+yAC,又AD=AB+BD,
∴AB+BD=xAB+yAC,所以BD=(x-1)AB+yAC,
又∵AC⊥BD,∴BD·AB=(x-1)AB²,
设AB=1,则DE=BC=√2,
又∵∠BED=60º,∴BD=√6/2,
显然BD与AB夹角45°,∴BD·AB=(x-1)AB²,√6/2×1×cos45º=(x-1)×12,x=√3/2+1,
同理,BD=(x-1)AB+yAC,两边同时乘以AC,由数量积得y=√3/2.