3. 模型规律:如图①,延长$CO$交$AB$于点$D$,则$∠BOC = ∠1 + ∠B = ∠A + ∠C + ∠B$.
(1)如图②,$∠A + ∠B + ∠C + ∠D + ∠E + ∠F =\_\_\_\_\_\_^{\circ}$;
(2)如图③,$∠ABO$,$∠ACO$的平分线$BO_1$,$CO_1$交于点$O_1$,已知$∠BOC = 120^{\circ}$,$∠A = 50^{\circ}$,则$∠BO_1C =\_\_\_\_\_\_^{\circ}$;
(3)如图④,$∠ABO$,$∠BAC$的平分线$BD$,$AD$交于点$D$,已知$∠BOC = 120^{\circ}$,$∠C = 44^{\circ}$,求$∠ADB$的度数.
(1)如图②,$∠A + ∠B + ∠C + ∠D + ∠E + ∠F =\_\_\_\_\_\_^{\circ}$;
(2)如图③,$∠ABO$,$∠ACO$的平分线$BO_1$,$CO_1$交于点$O_1$,已知$∠BOC = 120^{\circ}$,$∠A = 50^{\circ}$,则$∠BO_1C =\_\_\_\_\_\_^{\circ}$;
(3)如图④,$∠ABO$,$∠BAC$的平分线$BD$,$AD$交于点$D$,已知$∠BOC = 120^{\circ}$,$∠C = 44^{\circ}$,求$∠ADB$的度数.
答案：3.(1)260 解析:∠A+∠B+∠C+∠D+∠E+∠F=∠BOC+∠DOE=2×130°=260°.
(2)85 解析:∠BO₁C = ∠BOC - ∠OBO₁ - ∠OCO₁ = ∠BOC - $\frac{1}{2}$(∠ABO + ∠ACO) = ∠BOC - $\frac{1}{2}$(∠BOC - ∠A) = 120° - $\frac{1}{2}$×(120° - 50°) = 120° - 35° = 85°.
(3)∠ADB=180°−(∠ABD+∠BAD)=180°−$\frac{1}{2}$(∠ABO+∠BAC)=180°−$\frac{1}{2}$(∠BOC−∠C)=180°−$\frac{1}{2}$×(120°−44°)=142°.
(2)85 解析:∠BO₁C = ∠BOC - ∠OBO₁ - ∠OCO₁ = ∠BOC - $\frac{1}{2}$(∠ABO + ∠ACO) = ∠BOC - $\frac{1}{2}$(∠BOC - ∠A) = 120° - $\frac{1}{2}$×(120° - 50°) = 120° - 35° = 85°.
(3)∠ADB=180°−(∠ABD+∠BAD)=180°−$\frac{1}{2}$(∠ABO+∠BAC)=180°−$\frac{1}{2}$(∠BOC−∠C)=180°−$\frac{1}{2}$×(120°−44°)=142°.
4. (2025·菏泽期末)$△ ABC$是一个三角形的纸片,点$D$,$E$分别在边$AB$,$AC$上,将$∠A$沿$DE$折叠,点$A$落在点$A'$的位置.
(1)如图①,当点$A$落在边$AC$上时,若$∠BDA' = 62^{\circ}$,$∠A =$
(2)如图②,当点$A$落在$△ ABC$内部时,且$∠BDA' = 30^{\circ}$,$∠CEA' = 32^{\circ}$,求$∠A$的度数;
(3)如图③,当点$A$落在$△ ABC$外部且在$AC$下方时,猜想$∠BDA'$,$∠CEA'$和$∠A$的数量关系,并说明理由.
(1)如图①,当点$A$落在边$AC$上时,若$∠BDA' = 62^{\circ}$,$∠A =$
31°
;(2)如图②,当点$A$落在$△ ABC$内部时,且$∠BDA' = 30^{\circ}$,$∠CEA' = 32^{\circ}$,求$∠A$的度数;
(3)如图③,当点$A$落在$△ ABC$外部且在$AC$下方时,猜想$∠BDA'$,$∠CEA'$和$∠A$的数量关系,并说明理由.
答案：
4.(1)31° 解析:根据折叠的性质可知∠DA'E=∠A,
∵∠DA'E+∠A=∠BDA',
∴∠BDA'=2∠A=62°,即∠A=$\frac{1}{2}$×62°=31°.
(2)在四边形ADA'E中,∠A+∠DA'E+∠ADA'+∠A'EA=360°,
∴∠A+∠DA'E=360°−∠ADA'−∠A'EA.
∵∠BDA'+∠ADA'=180°,∠CEA'+∠A'EA=180°,
∴∠BDA'+∠CEA'=360°−∠ADA'−∠A'EA,
∴∠BDA'+∠CEA'=∠A+∠DA'E.
∵△A'DE是由△ADE沿直线DE折叠得到的,
∴∠A=∠DA'E,
∴∠BDA'+∠CEA'=2∠A.
∵∠BDA'=30°,∠CEA'=32°,
∴2∠A=62°,即∠A=$\frac{1}{2}$×62°=31°.
(3)∠BDA'−∠CEA'=2∠A.理由如下:
如图,DA'交AC于点F，
∵∠BDA'=∠A+∠DFA,∠DFA=∠A'+∠CEA',
∴∠BDA'=∠A+∠A'+∠CEA',
∴∠BDA'−∠CEA'=∠A+∠A'.
∵△A'DE是由△ADE沿直线DE折叠得到的，
∴∠A=∠A',
∴∠BDA'−∠CEA'=2∠A.
4.(1)31° 解析:根据折叠的性质可知∠DA'E=∠A,
∵∠DA'E+∠A=∠BDA',
∴∠BDA'=2∠A=62°,即∠A=$\frac{1}{2}$×62°=31°.
(2)在四边形ADA'E中,∠A+∠DA'E+∠ADA'+∠A'EA=360°,
∴∠A+∠DA'E=360°−∠ADA'−∠A'EA.
∵∠BDA'+∠ADA'=180°,∠CEA'+∠A'EA=180°,
∴∠BDA'+∠CEA'=360°−∠ADA'−∠A'EA,
∴∠BDA'+∠CEA'=∠A+∠DA'E.
∵△A'DE是由△ADE沿直线DE折叠得到的,
∴∠A=∠DA'E,
∴∠BDA'+∠CEA'=2∠A.
∵∠BDA'=30°,∠CEA'=32°,
∴2∠A=62°,即∠A=$\frac{1}{2}$×62°=31°.
(3)∠BDA'−∠CEA'=2∠A.理由如下:
如图,DA'交AC于点F，
∵∠BDA'=∠A+∠DFA,∠DFA=∠A'+∠CEA',
∴∠BDA'=∠A+∠A'+∠CEA',
∴∠BDA'−∠CEA'=∠A+∠A'.
∵△A'DE是由△ADE沿直线DE折叠得到的，
∴∠A=∠A',
∴∠BDA'−∠CEA'=2∠A.