∠BAD=α,∠CAD=β,则∠BAC=α+β
tanα=BD/AD=2/6=1/3
tanβ=CD/AD=3/6=1/2
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)
=(1/2+1/3)/(1-1/2*1/3)=1
又0