《2025年黄冈状元成才路状元大课堂七年级数学下册人教版》

5 - 1 (1)【问题解决】如图①，已知AB//CD，∠BEP = 36°，∠CFP = 152°，求∠EPF的度数.
(2)【问题迁移】如图②，若AB//CD，点P在直线AB的上方，则∠PFC，∠PEA，∠EPF之间有何数量关系？说明理由.
(3)【联想拓展】如图③，在(2)的条件下，已知∠EPF = α，∠PEA的平分线和∠PFC的平分线交于点G，求∠G的度数(结果用含α的式子表示).

答案：
5−1解:
(1)如图①,过点P作PG//AB,则∠EPG = ∠BEP = 36°.
∵AB//CD,
∴GP//CD,
∴∠FPG+∠CFP = 180°.
∴∠FPG = 180° - ∠CFP = 180° - 152° = 28°,
∴∠EPF = ∠EPG+∠FPG = 64°.

(2)∠EPF = ∠PFC - ∠PEA.理由如下:
如图②,过点P作PG//AB,
∴∠EPG = ∠PEA.
∵AB//CD,
∴PG//CD,
∴∠PFC = ∠FPG.
∴∠EPF = ∠FPG - ∠EPG = ∠PFC - ∠PEA.
(3)
∵FG平分∠PFC，EG平分∠PEA，
∴∠GFC = $\frac{1}{2}$∠PFC，∠GEA = $\frac{1}{2}$∠PEA.
由
(2)可得∠G = ∠GFC - ∠GEA.
∵∠EPF = ∠PFC - ∠PEA = α，
∴∠G = ∠GFC - ∠GEA = $\frac{1}{2}$∠PFC - $\frac{1}{2}$∠PEA = $\frac{1}{2}$(∠PFC - ∠PEA)= $\frac{1}{2}$α.

5 - 2 [广州增城区期中]已知AB//CD，∠ABE = 120°.
(1)如图①，求∠BED + ∠D的度数；
(2)如图②，∠DEF = 2∠BEF，∠CDF = 1/3∠CDE，EF与DF交于点F，求∠EFD的度数；
(3)如图③，过点B作BG⊥AB于点B，若∠CDE = 4∠GDE，求∠G/∠E的值.

答案：
5−2解:
(1)如图①,过点E作EF//AB,则∠ABE+∠FEB = 180°.
∵∠ABE = 120°,
∴∠FEB = 180° - ∠ABE = 60°.
∵AB//CD,
∴EF//CD,
∴∠FED+∠D = 180°.
∴∠FEB+∠BED+∠D = 180°.
∴∠BED+∠D = 180° - ∠FEB = 180° - 60° = 120°.

(2)设∠BEF = α，∠CDF = β.
∵∠DEF = 2∠BEF，∠CDF = $\frac{1}{3}$∠CDE，即∠CDE = 3∠CDF，
∴∠BED = 3α，∠CDE = 3β.
由
(1)知∠BED+∠CDE = 120°，
∴3α + 3β = 120°，
∴α + β = 40°.
如图②，过点E作EG//AB，过点F作FH//AB.
∵AB//CD，
∴EG//FH//CD，
∴∠GEF = ∠EFH，∠DFH = ∠CDF = β.
由
(1)知∠BEG = 60°，
∴∠EFD = ∠EFH+∠DFH = ∠GEF + ∠DFH = ∠BEG + ∠BEF+∠DFH = 60°+(α + β)= 60° + 40° = 100°.
(3)如图③，过点G作HG//AB，过点E作EI//AB.
∵AB//CD，
∴HG//AB//CD//EI.
∴∠HGD+∠CDG = 180°，∠IED+∠CDE = 180°.
∵∠CDE = 4∠GDE，
∴设∠GDE = x，则∠CDE = 4x，∠CDG = ∠CDE - ∠GDE = 3x.
∵BG⊥AB，
∴易得∠BGH = 90°.由
(1)可知∠BEI = 60°，
∴∠BGD = 180° - ∠BGH - ∠CDG = 90° - 3x，
∠BED = 180° - ∠BEI - ∠CDE = 120° - 4x.
∴$\frac{∠BGD}{∠BED}$ = $\frac{90° - 3x}{120° - 4x}$ = $\frac{3(30° - x)}{4(30° - x)}$ = $\frac{3}{4}$.