|AB+AC|^2=(AB+AC)·(AB+AC)=|AB|^2+|AC|^2+2AB·ACBC=AC-AB,故:|BC|^2=(AC-AB)·(AC-AB)=|AC|^2+|AB|^2-2AB·AC故:|AB|^2+|AC|^2+2AB·AC=|AC|^2+|AB|^2-2AB·AC,即:AB·AC=0即:AB⊥AC,故:|AC-AB|^2=(AC-AB)·(AC-AB)=|AC|^2+|AB|^2-2AB·AC=3+1=4,即:|AC-AB|=2