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1. （2025·连云港期末）如图①纸片$ABCD(AD// BC)$，将$CD$按如图②所示沿着$DE$折叠至$DC'$，$DC'$与线段$BC$交于点$F$，$∠ BFD=m$，点$E$在线段$BC$上，若将$AD$按如图③所示沿着$DO$折叠至$DA'$，且点$A'$在线段$DC$的延长线上，点$O$在线段$BC$上，则$∠ ODE=\_\_\_\_\_\_$.（用含$m$的式子表示）
答案
$90° - \frac{1}{2} m$
【解析】因为$∠BFD = m$,所以$∠DFC = 180° - m$. 因为$AD// BC$,所以$∠ADF = ∠DFC = 180° - m$, $∠C'DC = ∠ADC - (180° - m) = ∠ADC + m - 180°$. 由折叠性质得$∠CDE = \frac{1}{2}∠CDC' = \frac{1}{2}∠ADC + \frac{1}{2}m - 90°$, $∠CDO = ∠ADO = \frac{1}{2}∠ADC$, 所以$∠ODE = ∠CDO - ∠CDE = \frac{1}{2}∠ADC - (\frac{1}{2}∠ADC + \frac{1}{2}m - 90°) = 90° - \frac{1}{2}m$.
【解析】因为$∠BFD = m$,所以$∠DFC = 180° - m$. 因为$AD// BC$,所以$∠ADF = ∠DFC = 180° - m$, $∠C'DC = ∠ADC - (180° - m) = ∠ADC + m - 180°$. 由折叠性质得$∠CDE = \frac{1}{2}∠CDC' = \frac{1}{2}∠ADC + \frac{1}{2}m - 90°$, $∠CDO = ∠ADO = \frac{1}{2}∠ADC$, 所以$∠ODE = ∠CDO - ∠CDE = \frac{1}{2}∠ADC - (\frac{1}{2}∠ADC + \frac{1}{2}m - 90°) = 90° - \frac{1}{2}m$.
2. 如图,将长方形纸片ABCD沿EF折叠后,点C,D分别落在点C',D'的位置,C'D'交BC于点G,再将△C'FG沿FG折叠,点C'落在点C''的位置(点C''在折痕EF的左侧).
(1)如果∠FED'=65°,求∠EFC的度数;
(2)如果∠AED'=40°,则∠EFC''=
(3)探究∠EFC''与∠AED'的数量关系,并说明理由.
(1)如果∠FED'=65°,求∠EFC的度数;
(2)如果∠AED'=40°,则∠EFC''=
30
°;(3)探究∠EFC''与∠AED'的数量关系,并说明理由.
答案
(1) 根据题意,得$AD// BC$,所以$∠DEF + ∠EFC = 180°$. 因为折叠,且$∠FED' = 65°$,所以$∠DEF = ∠D'EF = 65°$,所以$∠EFC = 180° - ∠DEF = 115°$.
(2) 30 【解析】因为$∠AED' = 40°$,所以$∠DED' = 180° - ∠AED' = 140°$,由(1)知$∠DEF = ∠D'EF$,所以$∠DEF = ∠D'EF = \frac{1}{2}∠DED' = 70°$. 因为$AD// BC$,所以$∠EFC = 180° - ∠DEF = 110°$. 因为折叠,所以$∠EFC' = ∠EFC = 110°$. 因为$∠EFC = 110°$,所以$∠EFG = 180° - ∠EFC = 70°$,所以$∠C'FG = ∠EFC' - ∠EFG = 40°$. 因为折叠,所以$∠C'FG = ∠C''FG = 40°$,所以$∠EFC'' = ∠EFG - ∠GFC'' = 30°$.
(3) $∠EFC'' + \frac{3}{2}∠AED' = 90°$,理由:设$∠AED' = α$,所以$∠DED' = 180° - ∠AED' = 180° - α$. 由(1)知$∠DEF = ∠D'EF$,所以$∠DEF = ∠D'EF = \frac{1}{2}∠DED' = \frac{180° - α}{2}$. 因为$AD// BC$,所以$∠EFC = 180° - ∠DEF = 90° + \frac{1}{2}α$. 因为折叠,所以$∠EFC' = ∠EFC = 90° + \frac{1}{2}α$. 因为$∠EFC = 90° + \frac{1}{2}α$,所以$∠EFG = 180° - ∠EFC = 90° - \frac{1}{2}α$, $∠C'FG = ∠EFC' - ∠EFG = α$. 因为折叠,所以$∠C'FG = ∠C''FG = α$,所以$∠EFC'' = ∠EFG - ∠GFC'' = 90° - \frac{3}{2}α$, $∠EFC'' = 90° - \frac{3}{2}∠AED'$,所以$∠EFC'' + \frac{3}{2}∠AED' = 90°$.
(2) 30 【解析】因为$∠AED' = 40°$,所以$∠DED' = 180° - ∠AED' = 140°$,由(1)知$∠DEF = ∠D'EF$,所以$∠DEF = ∠D'EF = \frac{1}{2}∠DED' = 70°$. 因为$AD// BC$,所以$∠EFC = 180° - ∠DEF = 110°$. 因为折叠,所以$∠EFC' = ∠EFC = 110°$. 因为$∠EFC = 110°$,所以$∠EFG = 180° - ∠EFC = 70°$,所以$∠C'FG = ∠EFC' - ∠EFG = 40°$. 因为折叠,所以$∠C'FG = ∠C''FG = 40°$,所以$∠EFC'' = ∠EFG - ∠GFC'' = 30°$.
(3) $∠EFC'' + \frac{3}{2}∠AED' = 90°$,理由:设$∠AED' = α$,所以$∠DED' = 180° - ∠AED' = 180° - α$. 由(1)知$∠DEF = ∠D'EF$,所以$∠DEF = ∠D'EF = \frac{1}{2}∠DED' = \frac{180° - α}{2}$. 因为$AD// BC$,所以$∠EFC = 180° - ∠DEF = 90° + \frac{1}{2}α$. 因为折叠,所以$∠EFC' = ∠EFC = 90° + \frac{1}{2}α$. 因为$∠EFC = 90° + \frac{1}{2}α$,所以$∠EFG = 180° - ∠EFC = 90° - \frac{1}{2}α$, $∠C'FG = ∠EFC' - ∠EFG = α$. 因为折叠,所以$∠C'FG = ∠C''FG = α$,所以$∠EFC'' = ∠EFG - ∠GFC'' = 90° - \frac{3}{2}α$, $∠EFC'' = 90° - \frac{3}{2}∠AED'$,所以$∠EFC'' + \frac{3}{2}∠AED' = 90°$.
3. （2025·泰州月考）操作与探究：
如图，已知长方形的每个内角都是直角，将一张长方形纸片ABCD沿EF折叠后，点D，C的对应点分别是M，N，EM的延长线与BC相交于点G，MN与BC相交于点H.
(1)若$∠EFG=68°$，求$∠EGF$的度数.
(2)$∠AEM$与$∠BHM$的平分线相交于点Q，求$∠Q$的度数.
(3)设$∠FEM$与$∠GFE$的平分线相交于点P，$∠EFG$与$∠P$之间存在某种关系，你能找出这个关系吗？说明理由.
$\gg$ 进一步挑战进阶专题:P147 专题27
如图，已知长方形的每个内角都是直角，将一张长方形纸片ABCD沿EF折叠后，点D，C的对应点分别是M，N，EM的延长线与BC相交于点G，MN与BC相交于点H.
(1)若$∠EFG=68°$，求$∠EGF$的度数.
(2)$∠AEM$与$∠BHM$的平分线相交于点Q，求$∠Q$的度数.
(3)设$∠FEM$与$∠GFE$的平分线相交于点P，$∠EFG$与$∠P$之间存在某种关系，你能找出这个关系吗？说明理由.
$\gg$ 进一步挑战进阶专题:P147 专题27
答案
(1) 由题意得,$AD// BC$, $∠DEF = ∠GEF$,所以$∠DEF = ∠EFG = ∠GEF$. 因为$∠EFG = 68°$,所以$∠DEF = ∠GEF = 68°$, $∠AEG = 180° - (68° + 68°) = 44°$,所以$∠EGF = ∠AEG = 44°$.
(2) 如图,过点Q作$QR// AD$,所以$QR// AD// BC$, $∠EQR = ∠AEQ$, $∠HQR = ∠QHB$,所以$∠EQH = ∠AEQ + ∠QHB$. 同理可得,$∠EMH = ∠AEM + ∠MHB$. 因为$∠AEM$与$∠BHM$的平分线相交于点Q,所以$∠AEQ = \frac{1}{2}∠AEM$, $∠QHB = \frac{1}{2}∠MHB$,所以$∠EQH = \frac{1}{2}∠EMH = \frac{1}{2}×90° = 45°$.
(3) $∠P = 180° - ∠EFG$,理由如下:由(2)得,$∠P = ∠AEP + ∠PFB$. 因为$∠FEM$与$∠GFE$的平分线相交于点P,所以$∠PEG = \frac{1}{2}∠GEF = \frac{1}{2}∠FED = \frac{1}{2}∠EFG$, $∠PFG = \frac{1}{2}∠EFG$. 又因为$∠AEG = 180° - (∠FED + ∠GEF) = 180° - 2∠EFG$,所以$∠P = ∠AEP + ∠PFB = ∠AEG + ∠PEG + ∠PFB = (180° - 2∠EFG) + \frac{1}{2}∠EFG + \frac{1}{2}∠EFG = 180° - ∠EFG$.
