16. （1）如图①，小明把一副含$45^{\circ}$，$30^{\circ}$角的直角三角尺按下图摆放，其中$∠ C = ∠ F = 90^{\circ}$，$∠ A = 45^{\circ}$，$∠ D = 30^{\circ}$，求$∠ α + ∠ β$的度数；
（2）如图②，已知$∠ BOF = 120^{\circ}$，试求$∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F$的度数。
（2）如图②，已知$∠ BOF = 120^{\circ}$，试求$∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F$的度数。
答案：
16. (1)如图①,由三角形外角的性质,得∠α = ∠1 + ∠D,∠β = ∠4 + ∠F,所以∠α + ∠β = ∠1 + ∠D + ∠4 + ∠F.又∠1 = ∠2,∠3 = ∠4,∠2 + ∠3 = 90°,所以∠α + ∠β = ∠2 + ∠D + ∠3 + ∠F = ∠2 + ∠3 + 30° + 90° = 210°.
(2)如图②,由三角形的外角性质,得∠1 = ∠A + ∠C,∠2 = ∠B + ∠D,因为∠BOF = 120°,所以∠3 = 180° - 120° = 60°.由三角形内角和定理,得∠E + ∠1 = 180° - 60° = 120°,∠F + ∠2 = 180° - 60° = 120°,所以∠1 + ∠2 + ∠E + ∠F = 120° + 120° = 240°,即∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 240°.
16. (1)如图①,由三角形外角的性质,得∠α = ∠1 + ∠D,∠β = ∠4 + ∠F,所以∠α + ∠β = ∠1 + ∠D + ∠4 + ∠F.又∠1 = ∠2,∠3 = ∠4,∠2 + ∠3 = 90°,所以∠α + ∠β = ∠2 + ∠D + ∠3 + ∠F = ∠2 + ∠3 + 30° + 90° = 210°.
(2)如图②,由三角形的外角性质,得∠1 = ∠A + ∠C,∠2 = ∠B + ∠D,因为∠BOF = 120°,所以∠3 = 180° - 120° = 60°.由三角形内角和定理,得∠E + ∠1 = 180° - 60° = 120°,∠F + ∠2 = 180° - 60° = 120°,所以∠1 + ∠2 + ∠E + ∠F = 120° + 120° = 240°,即∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 240°.
17. 一题多变 (1)如图①，将$△ ABC$纸片沿$DE$进行折叠，使点$A$落在四边形$BCED$的外部点$A'$的位置，若$∠ A = 35^{\circ}$，则$∠ 1 - ∠ 2 =\_\_\_\_\_\_^{\circ}$；
(2)(2025·南京期末)如图②，将$△ ABC$纸片先沿$DE$折叠，再沿$FG$折叠，若$∠ 1 + ∠ 2 = 228^{\circ}$，则$∠ 3 + ∠ 4 =\_\_\_\_\_\_^{\circ}$。
(2)(2025·南京期末)如图②，将$△ ABC$纸片先沿$DE$折叠，再沿$FG$折叠，若$∠ 1 + ∠ 2 = 228^{\circ}$，则$∠ 3 + ∠ 4 =\_\_\_\_\_\_^{\circ}$。
答案：
17. (1)70 解析:如图①所示,
∵△ABC纸片沿DE进行折叠,点A落在四边形BCED的外部点A'的位置,
∴∠4 = ∠5,∠3 = ∠2 + ∠DEC.
∵∠1 + ∠4 + ∠5 = 180°,
∴∠1 + 2∠4 = 180°,
∴∠1 = 180° - 2∠4.
∵∠3 + ∠DEC = 180°,
∴∠2 = ∠3 - ∠DEC = 2∠3 - 180°,
∴∠1 - ∠2 = 180° - 2∠4 - 2∠3 + 180° = 360° - 2∠4 - 2∠3 = 2∠A,
∴∠1 - ∠2 = 2×35° = 70°.
(2)96 解析:如图②,
∵∠1 + ∠2 = 228°,∠1 = ∠A' + ∠A'NM,∠2 = ∠A' + ∠A'MN,
∴2∠A' + ∠A'NM + ∠A'MN = 228°.
∵∠A' + ∠A'NM + ∠A'MN = 180°,
∴∠A' = 228° - 180° = 48°,
∴∠A = ∠A' = 48°,
∴∠AED + ∠ADE = 180° - 48° = 132°,
∴∠AEF + ∠ADG = 2(∠AED + ∠ADE)=2×132° = 264°,
∴∠3 + ∠4 = 360° - 264° = 96°.
17. (1)70 解析:如图①所示,
∵△ABC纸片沿DE进行折叠,点A落在四边形BCED的外部点A'的位置,
∴∠4 = ∠5,∠3 = ∠2 + ∠DEC.
∵∠1 + ∠4 + ∠5 = 180°,
∴∠1 + 2∠4 = 180°,
∴∠1 = 180° - 2∠4.
∵∠3 + ∠DEC = 180°,
∴∠2 = ∠3 - ∠DEC = 2∠3 - 180°,
∴∠1 - ∠2 = 180° - 2∠4 - 2∠3 + 180° = 360° - 2∠4 - 2∠3 = 2∠A,
∴∠1 - ∠2 = 2×35° = 70°.
(2)96 解析:如图②,
∵∠1 + ∠2 = 228°,∠1 = ∠A' + ∠A'NM,∠2 = ∠A' + ∠A'MN,
∴2∠A' + ∠A'NM + ∠A'MN = 228°.
∵∠A' + ∠A'NM + ∠A'MN = 180°,
∴∠A' = 228° - 180° = 48°,
∴∠A = ∠A' = 48°,
∴∠AED + ∠ADE = 180° - 48° = 132°,
∴∠AEF + ∠ADG = 2(∠AED + ∠ADE)=2×132° = 264°,
∴∠3 + ∠4 = 360° - 264° = 96°.
18. 在$△ ABC$和$△ ACD$(共$AC$边且不重合)中，$∠ B = ∠ BAC$，$∠ D = ∠ DAC$。
(1)如图①，当$△ ABC$和$△ ACD$均为钝角三角形，点$B$，$D$在直线$AC$两侧时，$∠ BCD$和$∠ BAD$之间的数量关系为
(2)如图②，当$△ ABC$和$△ ACD$均为锐角三角形，且点$B$，$D$在直线$AC$两侧时，$∠ BCD$和$∠ BAD$之间的数量关系为
(3)如图③，当$△ ABC$为钝角三角形，$△ ACD$为锐角三角形，且$B$，$D$在直线$AC$同侧时，求证：$∠ BCD = 2∠ BAD$；
(4)分别作$∠ B$和$∠ D$的平分线，两条角平分线所在直线交点于点$P$(点$P$不与点$B$或者点$D$重合)，当$∠ BCD = 100^{\circ}$时，直接写出$∠ BPD$的度数。
(1)如图①，当$△ ABC$和$△ ACD$均为钝角三角形，点$B$，$D$在直线$AC$两侧时，$∠ BCD$和$∠ BAD$之间的数量关系为
∠BCD = 2∠BAD
；(2)如图②，当$△ ABC$和$△ ACD$均为锐角三角形，且点$B$，$D$在直线$AC$两侧时，$∠ BCD$和$∠ BAD$之间的数量关系为
2∠BAD + ∠BCD = 360°
；(3)如图③，当$△ ABC$为钝角三角形，$△ ACD$为锐角三角形，且$B$，$D$在直线$AC$同侧时，求证：$∠ BCD = 2∠ BAD$；
(4)分别作$∠ B$和$∠ D$的平分线，两条角平分线所在直线交点于点$P$(点$P$不与点$B$或者点$D$重合)，当$∠ BCD = 100^{\circ}$时，直接写出$∠ BPD$的度数。
答案：
18. (1)∠BCD = 2∠BAD 解析:如图①所示,延长AC至点E,在△ABC,△ADC中,∠1 = ∠B + ∠BAC,∠2 = ∠D + ∠DAC.
∵∠B = ∠BAC,∠D = ∠DAC,
∴∠1 + ∠2 = (∠B + ∠BAC)+(∠D + ∠DAC)=2∠BAC + 2∠DAC = 2(∠BAC + ∠DAC)=2∠BAD,
∴∠BCD = 2∠BAD.
(2)2∠BAD + ∠BCD = 360° 解析:由△ABC和△ACD两个三角形组成了四边形ABCD,
∴∠B + ∠BAC + ∠DAC + ∠D + ∠DCA + ∠ACB = 360°.
∵∠B = ∠BAC,∠D = ∠DAC,
∴2∠BAC + 2∠DAC + ∠DCA + ∠ACB = 360°,
∴2(∠BAC + ∠DAC)+(∠DCA + ∠ACB)=360°,
∴2∠BAD + ∠BCD = 360°.
(3)如图②所示,设AB,CD交于点E,在△BCE中,∠BED = ∠B + ∠BCD,在△ADE中,∠BED = ∠BAD + ∠D,
∴∠B + ∠BCD = ∠BAD + ∠D.又
∵∠B = ∠BAC,∠D = ∠DAC,
∴∠BCD = ∠BAD + ∠D - ∠B = ∠BAD+(∠DAC - ∠BAC).
∵∠DAC - ∠BAC = ∠BAD,
∴∠BCD = ∠BAD + ∠BAD,
∴∠BCD = 2∠BAD.
(4)∠BPD的度数为75°或165°或15°. 解析:①如图③所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,
∴∠ABP = ∠PBC = $\frac{1}{2}$∠ABC = $\frac{1}{2}$∠BAC,∠ADP = ∠PDC = $\frac{1}{2}$∠ADC = $\frac{1}{2}$∠DAC.
∵∠BPC = ∠BAP + ∠ABP = $\frac{3}{2}$∠BAC,∠DPC = ∠DAP + ∠PDA = $\frac{3}{2}$∠DAC,
∴∠BPC + ∠DPC = $\frac{3}{2}$(∠BAC + ∠DAC)= $\frac{3}{2}$∠BAD,
∴∠BPD = $\frac{3}{2}$∠BAD.由(1)可知,∠BCD = 2∠BAD,
∴∠BAD = $\frac{1}{2}$∠BCD = $\frac{1}{2}$×100° = 50°,
∴∠BPD = $\frac{3}{2}$∠BAD = $\frac{3}{2}$×50° = 75°.
②如图④所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,且∠CBA = ∠BAC,∠ADC = ∠DAC,
∴∠CBP = ∠PBA = $\frac{1}{2}$∠ABC = $\frac{1}{2}$∠BAC,∠CDP = ∠PDA = $\frac{1}{2}$∠ADC = $\frac{1}{2}$∠DAC,
∴∠PBA + ∠PDA = $\frac{1}{2}$(∠BAC + ∠DAC)= $\frac{1}{2}$∠BAD.由(2)可知,∠BCD + 2∠BAD = 360°,
∴∠BAD = $\frac{360° - ∠BCD}{2}$ = $\frac{360° - 100°}{2}$ = 130°.在四边形ABPD中,连接AP,可得两个三角形的内角和等于四边形ABPD的内角和,
∴∠BPD = 180° + 180° - ∠BAD - (∠PBA + ∠PDA)=360° - $\frac{3}{2}$∠BAD = 360° - $\frac{3}{2}$×130° = 165°.
③如图⑤所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,设AB,DP交于点F,AB,CD交于点E,在△ADF中,∠BAD + $\frac{1}{2}$∠CDA = ∠BFD,在△BPF中,$\frac{1}{2}$∠CBE + ∠P = ∠BFD,
∴∠BAD + $\frac{1}{2}$∠CDA = $\frac{1}{2}$∠CBE + ∠P,
∴∠P = ∠BAD + $\frac{1}{2}$∠CDA - $\frac{1}{2}$∠CBE = ∠BAD + $\frac{1}{2}$(∠CDA - ∠CBE).
∵∠CBA = ∠BAC,∠ADC = ∠DAC,
∴∠P = ∠BAD + $\frac{1}{2}$(∠CAD - ∠CAB)=∠BAD + $\frac{1}{2}$∠BAD = $\frac{3}{2}$∠BAD.由(3)可知,∠BCD = 2∠BAD,
∴∠BAD = $\frac{1}{2}$∠BCD = $\frac{1}{2}$×100° = 50°,
∴∠P = $\frac{3}{2}$∠BAD = $\frac{3}{2}$×50° = 75°.
④如图⑥所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,设直线BP交AC于M,直线DP交AC于N.
∵∠BCD = 100°,
∴∠CBA + ∠CAB + ∠CDA + ∠CAD = 360° - 100° = 260°.
∵∠CBA = ∠CAB,∠CDA = ∠CAD,
∴∠CBA + ∠CDA = $\frac{1}{2}$×260° = 130°.
∵∠PMN = ∠CBM + ∠BCM,∠PNM = ∠NCD + ∠NDC,
∴∠PMN + ∠PNM = $\frac{1}{2}$∠CBA + ∠BCM + ∠NCD + $\frac{1}{2}$∠CDA = $\frac{1}{2}$(∠CBA + ∠CDA)+∠BCD = $\frac{1}{2}$×130° + 100° = 165°,
∴∠BPD = 180° - 165° = 15°.综上所示,∠BPD的度数为75°或165°或15°.
18. (1)∠BCD = 2∠BAD 解析:如图①所示,延长AC至点E,在△ABC,△ADC中,∠1 = ∠B + ∠BAC,∠2 = ∠D + ∠DAC.
∵∠B = ∠BAC,∠D = ∠DAC,
∴∠1 + ∠2 = (∠B + ∠BAC)+(∠D + ∠DAC)=2∠BAC + 2∠DAC = 2(∠BAC + ∠DAC)=2∠BAD,
∴∠BCD = 2∠BAD.
(2)2∠BAD + ∠BCD = 360° 解析:由△ABC和△ACD两个三角形组成了四边形ABCD,
∴∠B + ∠BAC + ∠DAC + ∠D + ∠DCA + ∠ACB = 360°.
∵∠B = ∠BAC,∠D = ∠DAC,
∴2∠BAC + 2∠DAC + ∠DCA + ∠ACB = 360°,
∴2(∠BAC + ∠DAC)+(∠DCA + ∠ACB)=360°,
∴2∠BAD + ∠BCD = 360°.
(3)如图②所示,设AB,CD交于点E,在△BCE中,∠BED = ∠B + ∠BCD,在△ADE中,∠BED = ∠BAD + ∠D,
∴∠B + ∠BCD = ∠BAD + ∠D.又
∵∠B = ∠BAC,∠D = ∠DAC,
∴∠BCD = ∠BAD + ∠D - ∠B = ∠BAD+(∠DAC - ∠BAC).
∵∠DAC - ∠BAC = ∠BAD,
∴∠BCD = ∠BAD + ∠BAD,
∴∠BCD = 2∠BAD.
(4)∠BPD的度数为75°或165°或15°. 解析:①如图③所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,
∴∠ABP = ∠PBC = $\frac{1}{2}$∠ABC = $\frac{1}{2}$∠BAC,∠ADP = ∠PDC = $\frac{1}{2}$∠ADC = $\frac{1}{2}$∠DAC.
∵∠BPC = ∠BAP + ∠ABP = $\frac{3}{2}$∠BAC,∠DPC = ∠DAP + ∠PDA = $\frac{3}{2}$∠DAC,
∴∠BPC + ∠DPC = $\frac{3}{2}$(∠BAC + ∠DAC)= $\frac{3}{2}$∠BAD,
∴∠BPD = $\frac{3}{2}$∠BAD.由(1)可知,∠BCD = 2∠BAD,
∴∠BAD = $\frac{1}{2}$∠BCD = $\frac{1}{2}$×100° = 50°,
∴∠BPD = $\frac{3}{2}$∠BAD = $\frac{3}{2}$×50° = 75°.
②如图④所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,且∠CBA = ∠BAC,∠ADC = ∠DAC,
∴∠CBP = ∠PBA = $\frac{1}{2}$∠ABC = $\frac{1}{2}$∠BAC,∠CDP = ∠PDA = $\frac{1}{2}$∠ADC = $\frac{1}{2}$∠DAC,
∴∠PBA + ∠PDA = $\frac{1}{2}$(∠BAC + ∠DAC)= $\frac{1}{2}$∠BAD.由(2)可知,∠BCD + 2∠BAD = 360°,
∴∠BAD = $\frac{360° - ∠BCD}{2}$ = $\frac{360° - 100°}{2}$ = 130°.在四边形ABPD中,连接AP,可得两个三角形的内角和等于四边形ABPD的内角和,
∴∠BPD = 180° + 180° - ∠BAD - (∠PBA + ∠PDA)=360° - $\frac{3}{2}$∠BAD = 360° - $\frac{3}{2}$×130° = 165°.
③如图⑤所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,设AB,DP交于点F,AB,CD交于点E,在△ADF中,∠BAD + $\frac{1}{2}$∠CDA = ∠BFD,在△BPF中,$\frac{1}{2}$∠CBE + ∠P = ∠BFD,
∴∠BAD + $\frac{1}{2}$∠CDA = $\frac{1}{2}$∠CBE + ∠P,
∴∠P = ∠BAD + $\frac{1}{2}$∠CDA - $\frac{1}{2}$∠CBE = ∠BAD + $\frac{1}{2}$(∠CDA - ∠CBE).
∵∠CBA = ∠BAC,∠ADC = ∠DAC,
∴∠P = ∠BAD + $\frac{1}{2}$(∠CAD - ∠CAB)=∠BAD + $\frac{1}{2}$∠BAD = $\frac{3}{2}$∠BAD.由(3)可知,∠BCD = 2∠BAD,
∴∠BAD = $\frac{1}{2}$∠BCD = $\frac{1}{2}$×100° = 50°,
∴∠P = $\frac{3}{2}$∠BAD = $\frac{3}{2}$×50° = 75°.
④如图⑥所示,BP,DP分别是∠ABC,∠ADC的平分线,∠BCD = 100°,设直线BP交AC于M,直线DP交AC于N.
∵∠BCD = 100°,
∴∠CBA + ∠CAB + ∠CDA + ∠CAD = 360° - 100° = 260°.
∵∠CBA = ∠CAB,∠CDA = ∠CAD,
∴∠CBA + ∠CDA = $\frac{1}{2}$×260° = 130°.
∵∠PMN = ∠CBM + ∠BCM,∠PNM = ∠NCD + ∠NDC,
∴∠PMN + ∠PNM = $\frac{1}{2}$∠CBA + ∠BCM + ∠NCD + $\frac{1}{2}$∠CDA = $\frac{1}{2}$(∠CBA + ∠CDA)+∠BCD = $\frac{1}{2}$×130° + 100° = 165°,
∴∠BPD = 180° - 165° = 15°.综上所示,∠BPD的度数为75°或165°或15°.