模型1如图1,有结论:∠BOC=∠A+∠B+∠C.证明:如图2,连接A0并延长到E.∠BOE=∠BAO+∠B.∠COE=∠CAO+∠C.∴∠BOE+∠COE=∠BAO+∠B+∠CAO+∠C,即∠BOC=∠BAC+∠B+∠C.模型2如图3,当BO,CO分别平分∠ABC与∠ACB时,则有∠BOC=90°+1/2∠A.证明:∠BOC=180°-(∠OBC+∠OCB)=180°-(1/2∠ABC+1/2∠ACB)=180°-1/2(∠ABC+∠ACB) Model 1, as shown in Figure 1, concludes: ∠BOC = ∠A + ∠B + ∠C. Proof: As shown in Figure 2, connect A0 and extend to E.∠BOE = ∠BAO + ∠B.∠COE = ∠CAO + ∠C.∴∠ BOE + ∠COE = ∠BAO + ∠B + ∠CAO + ∠C, ie ∠BOC = ∠BAC + ∠B + ∠C. Model 2 As shown in Figure 3, when BO and CO are divided by ∠ABC and ∠ACB respectively, then ∠BOC = 90 ° + 1 / 2∠A. Proof: ∠BOC = 180 ° - (∠OBC + ∠OCB) = 180 ° - (1/2 ∠ABC + 1 / 2∠ACB) = 180 ° -1/2 (∠ABC + ∠ACB )