3. (2025·南通期末)折纸中的数学:
已知,在长方形纸片ABCD中,∠A=∠B=∠C=∠D,AD//BC,AB//CD.点M在边AD上,点E在边CD上,将长方形ABCD按如图①方式沿ME折叠,点D的对应点为D',连接MD'并延长,交BC于点N,设∠MNC=α(45°＜α＜90°).
【初步探索】(1)若α=70°,则∠MED=
【深入探究】(2)如图②,将四边形MNCD沿MN翻折至四边形MNPG,PG交AD于点H,写出∠PNB和∠GMH的数量关系:
【拓展延伸】(3)如图③,在(2)的条件下,连接HN,将△PNH沿HN翻折至△P'NH,当HP平分∠AHN时,求∠PNB和∠P'NH的数量关系,用含α的式子直接表示出∠P'NM的大小,并写出α的范围.
已知,在长方形纸片ABCD中,∠A=∠B=∠C=∠D,AD//BC,AB//CD.点M在边AD上,点E在边CD上,将长方形ABCD按如图①方式沿ME折叠,点D的对应点为D',连接MD'并延长,交BC于点N,设∠MNC=α(45°＜α＜90°).
【初步探索】(1)若α=70°,则∠MED=
35
°;【深入探究】(2)如图②,将四边形MNCD沿MN翻折至四边形MNPG,PG交AD于点H,写出∠PNB和∠GMH的数量关系:
∠PNB = ∠GMH
;【拓展延伸】(3)如图③,在(2)的条件下,连接HN,将△PNH沿HN翻折至△P'NH,当HP平分∠AHN时,求∠PNB和∠P'NH的数量关系,用含α的式子直接表示出∠P'NM的大小,并写出α的范围.
答案：(1)35 解析:
∵∠MNC = α,AD//BC,
∴∠DMN = 180° - α,∠C + ∠D = 180°.又
∵∠C = ∠D,
∴∠C = ∠D = 90°,由折叠性质可得∠DME = ∠D'ME = $\frac{1}{2}$∠DMN = 90° - $\frac{1}{2}$α,∠D = ∠MD'E = 90°,
∴∠DED' = 180° - ∠DMD' = α,
∴∠MED = $\frac{1}{2}$α.
∵α = 70°,
∴∠MED = 35°.
(2)∠PNB = ∠GMH 解析:由折叠的性质可得∠PNM = ∠CNM = α,
∴∠PNB = 180° - ∠PNM - ∠CNM = 180° - 2α.
∵AD//BC,
∴∠DMN = 180° - ∠MNC = 180° - α,∠HMN = ∠MNC = α,由折叠的性质可得∠GMN = ∠DMN = 180° - α,
∴∠GMH = ∠GMN - ∠HMN = 180° - α - α = 180° - 2α,
∴∠PNB = ∠GMH.
(3)由(2)可得∠PNB = ∠GMH = 180° - 2α,由折叠可得∠G = ∠D = 90°,∠C = ∠P = 90°,
∴∠AHP = ∠GMH = 90° - ∠GMH = 90° - (180° - 2α) = 2α - 90°.
∵△PNH沿HN翻折至△P'NH,
∴∠P'NH = ∠PNH = 90° - ∠PHN = 90° - ∠AHP = 90° - (2α - 90°) = 180° - 2α,
∴∠PNB = ∠P'NH.
∵∠PNM = ∠CNM = α,
∴∠HNM = ∠PNM - ∠PNH = α - (180° - 2α) = 3α - 180°,当180° - 2α＞3α - 180°,即∠HNP'＞∠MNH时,解得α＜72°,
∴∠P'NM = ∠P'NH - ∠HNM = 180° - 2α - (3α - 180°) = 360° - 5α.
当180° - 2α＜3α - 180°,即∠P'NH＜∠HNM时,α＞72°,
∴∠P'NM = ∠HNM - ∠P'NH = 5α - 360°.
综上所述,∠PNB = ∠P'NH;∠P'NM = 5α - 360°(α＞72°)或∠P'NM = 360° - 5α(α＜72°).
∵∠MNC = α,AD//BC,
∴∠DMN = 180° - α,∠C + ∠D = 180°.又
∵∠C = ∠D,
∴∠C = ∠D = 90°,由折叠性质可得∠DME = ∠D'ME = $\frac{1}{2}$∠DMN = 90° - $\frac{1}{2}$α,∠D = ∠MD'E = 90°,
∴∠DED' = 180° - ∠DMD' = α,
∴∠MED = $\frac{1}{2}$α.
∵α = 70°,
∴∠MED = 35°.
(2)∠PNB = ∠GMH 解析:由折叠的性质可得∠PNM = ∠CNM = α,
∴∠PNB = 180° - ∠PNM - ∠CNM = 180° - 2α.
∵AD//BC,
∴∠DMN = 180° - ∠MNC = 180° - α,∠HMN = ∠MNC = α,由折叠的性质可得∠GMN = ∠DMN = 180° - α,
∴∠GMH = ∠GMN - ∠HMN = 180° - α - α = 180° - 2α,
∴∠PNB = ∠GMH.
(3)由(2)可得∠PNB = ∠GMH = 180° - 2α,由折叠可得∠G = ∠D = 90°,∠C = ∠P = 90°,
∴∠AHP = ∠GMH = 90° - ∠GMH = 90° - (180° - 2α) = 2α - 90°.
∵△PNH沿HN翻折至△P'NH,
∴∠P'NH = ∠PNH = 90° - ∠PHN = 90° - ∠AHP = 90° - (2α - 90°) = 180° - 2α,
∴∠PNB = ∠P'NH.
∵∠PNM = ∠CNM = α,
∴∠HNM = ∠PNM - ∠PNH = α - (180° - 2α) = 3α - 180°,当180° - 2α＞3α - 180°,即∠HNP'＞∠MNH时,解得α＜72°,
∴∠P'NM = ∠P'NH - ∠HNM = 180° - 2α - (3α - 180°) = 360° - 5α.
当180° - 2α＜3α - 180°,即∠P'NH＜∠HNM时,α＞72°,
∴∠P'NM = ∠HNM - ∠P'NH = 5α - 360°.
综上所述,∠PNB = ∠P'NH;∠P'NM = 5α - 360°(α＞72°)或∠P'NM = 360° - 5α(α＜72°).
4. (2025·常州期中)如图,已知MN//GH,点C在MN上,点A,B在GH上.在△ABC中,∠ACB=90°,∠BAC=45°,点E,F在直线BC上,在△DEF中,∠EDF=90°,∠DFE=30°.
(1)图中∠BCN的度数是多少?请说明理由.
(2)将△DEF沿直线BC平移,使得点E与B重合,再将△DEF绕点E按逆时针方向进行旋转,至少旋转
(3)将△DEF沿直线BC平移,当点D在MN上时,求∠CDE的度数.
(4)将△DEF沿直线BC平移,当以C,D,F为顶点的三角形中有两个角相等时,请直接写出∠CDE的度数.
(1)图中∠BCN的度数是多少?请说明理由.
(2)将△DEF沿直线BC平移,使得点E与B重合,再将△DEF绕点E按逆时针方向进行旋转,至少旋转
30
度,使得DE与AC平行.(3)将△DEF沿直线BC平移,当点D在MN上时,求∠CDE的度数.
(4)将△DEF沿直线BC平移,当以C,D,F为顶点的三角形中有两个角相等时,请直接写出∠CDE的度数.
答案：
(1)∠BCN = 45°,理由如下:
∵∠ACB + ∠ABC + ∠BAC = 180°,∠ACB = 90°,∠BAC = 45°,
∴∠ABC = 45°.
∵MN//GH,
∴∠BCN = ∠ABC = 45°.
(2)30 解析:如图①,
∵D'E//AC,
∴∠D'EH = ∠BAC = 45°.
∵∠ABC = 45°,
∴∠HEF = 45°.
∵∠EDF = 90°,∠DFE = 30°,
∴∠DEF = 180° - ∠EDF - ∠DFE = 60°,
∴∠DEH = ∠DEF - ∠HEF = 15°,
∴∠D'ED = ∠D'EH - ∠DEH = 30°,
∴将△DEF绕点E按逆时针方向进行旋转，至少旋转30°,使得DE//AC.
(3)如图②,
∵∠ACB + ∠ABC + ∠BAC = 180°,∠ACB = 90°,∠BAC = 45°,
∴∠ABC = 45°.
∵MN//GH,
∴∠DCE = ∠ABC = 45°.
∵在△DEF中,∠EDF = 90°,∠DFE = 30°,
∴∠DEF = 180° - ∠EDF - ∠DFE = 60°.
∵∠CED + ∠DEF = 180°,
∴∠CED = 120°.
∵∠DCE + ∠CDE + ∠CED = 180°,
∴∠CDE = 180° - ∠DCE - ∠CED = 15°.
(4)∠CDE = 60°或105°或15°或30°.解析:分两种情况，I.当△DEF向上平移时,①如图③所示,当以C,D,F为顶点的三角形中有两个角相等,即∠DFE = ∠CDF = 30°时,
∵∠EDF = ∠CDE + ∠CDF = 90°,
∴∠CDE = 90° - ∠CDF = 60°.
②如图④所示，当以C,D,F为顶点的三角形中有两个角相等,即∠FDC = ∠DCF时,
∵∠DFE = ∠FDC + ∠DCF = 30°,
∴∠FDC = ∠DCF = 15°.
∵∠EDF = 90°,
∴∠CDE = ∠EDF + ∠FDC = 90° + 15° = 105°.
③如图⑤所示，当以C,D,F为顶点的三角形中有两个角相等,即∠FDC = ∠DCF时,
∵∠DFE = 30°,∠FDC + ∠DCF + ∠DFE = 180°,
∴∠FDC = ∠DCF = 75°.
∵∠EDF = 90°,
∴∠CDE = ∠EDF - ∠FDC = 90° - 75° = 15°.
II.当△DEF向下平移时,如图⑥所示，
④当以C,D,F为顶点的三角形中有两个角相等，即∠DFE = ∠DCF = 30°时,
∵∠EDF = 90°,
∴∠CED = ∠EDF + ∠DFE = 120°,
∴∠CDE = 180° - ∠CED - ∠DCF = 30°.
综上可知,将△DEF沿直线BC平移,当以C,D,F为顶点的三角形中有两个角相等时∠CDE的度数为60°或105°或15°或30°.
(1)∠BCN = 45°,理由如下:
∵∠ACB + ∠ABC + ∠BAC = 180°,∠ACB = 90°,∠BAC = 45°,
∴∠ABC = 45°.
∵MN//GH,
∴∠BCN = ∠ABC = 45°.
(2)30 解析:如图①,
∵D'E//AC,
∴∠D'EH = ∠BAC = 45°.
∵∠ABC = 45°,
∴∠HEF = 45°.
∵∠EDF = 90°,∠DFE = 30°,
∴∠DEF = 180° - ∠EDF - ∠DFE = 60°,
∴∠DEH = ∠DEF - ∠HEF = 15°,
∴∠D'ED = ∠D'EH - ∠DEH = 30°,
∴将△DEF绕点E按逆时针方向进行旋转，至少旋转30°,使得DE//AC.
(3)如图②,
∵∠ACB + ∠ABC + ∠BAC = 180°,∠ACB = 90°,∠BAC = 45°,
∴∠ABC = 45°.
∵MN//GH,
∴∠DCE = ∠ABC = 45°.
∵在△DEF中,∠EDF = 90°,∠DFE = 30°,
∴∠DEF = 180° - ∠EDF - ∠DFE = 60°.
∵∠CED + ∠DEF = 180°,
∴∠CED = 120°.
∵∠DCE + ∠CDE + ∠CED = 180°,
∴∠CDE = 180° - ∠DCE - ∠CED = 15°.
(4)∠CDE = 60°或105°或15°或30°.解析:分两种情况，I.当△DEF向上平移时,①如图③所示,当以C,D,F为顶点的三角形中有两个角相等,即∠DFE = ∠CDF = 30°时,
∵∠EDF = ∠CDE + ∠CDF = 90°,
∴∠CDE = 90° - ∠CDF = 60°.
②如图④所示，当以C,D,F为顶点的三角形中有两个角相等,即∠FDC = ∠DCF时,
∵∠DFE = ∠FDC + ∠DCF = 30°,
∴∠FDC = ∠DCF = 15°.
∵∠EDF = 90°,
∴∠CDE = ∠EDF + ∠FDC = 90° + 15° = 105°.
③如图⑤所示，当以C,D,F为顶点的三角形中有两个角相等,即∠FDC = ∠DCF时,
∵∠DFE = 30°,∠FDC + ∠DCF + ∠DFE = 180°,
∴∠FDC = ∠DCF = 75°.
∵∠EDF = 90°,
∴∠CDE = ∠EDF - ∠FDC = 90° - 75° = 15°.
II.当△DEF向下平移时,如图⑥所示，④当以C,D,F为顶点的三角形中有两个角相等，即∠DFE = ∠DCF = 30°时,
∵∠EDF = 90°,
∴∠CED = ∠EDF + ∠DFE = 120°,
∴∠CDE = 180° - ∠CED - ∠DCF = 30°.
综上可知,将△DEF沿直线BC平移,当以C,D,F为顶点的三角形中有两个角相等时∠CDE的度数为60°或105°或15°或30°.