1.(2024·如皋期中)完成下面的证明:
如图,AD⊥BC,FG⊥BC,垂足分别为D,G,且∠GFC = ∠ADE.
求证:DE//AC.
证明:∵ AD⊥BC,FG⊥BC(已知),
∴ ∠ADC = ________°,∠FGC = ________°(垂直的定义).
∴ ∠ADC = ∠FGC.
∴ AD//________(同位角相等,两直线平行).
∴ ∠DAC = ∠GFC( ).
又∵ ∠GFC = ∠ADE(已知),∴ ∠DAC = ∠ADE. ∴ DE//AC( ).

如图,AD⊥BC,FG⊥BC,垂足分别为D,G,且∠GFC = ∠ADE.
求证:DE//AC.
证明:∵ AD⊥BC,FG⊥BC(已知),
∴ ∠ADC = ________°,∠FGC = ________°(垂直的定义).
∴ ∠ADC = ∠FGC.
∴ AD//________(同位角相等,两直线平行).
∴ ∠DAC = ∠GFC( ).
又∵ ∠GFC = ∠ADE(已知),∴ ∠DAC = ∠ADE. ∴ DE//AC( ).
答案
90 90 FG 两直线平行,同位角相等 内错角相等,两直线平行
2. 如图,AB//CD,∠B = ∠D,直线EF与AD,BC的延长线分别交于点E,F. 求证:∠DEF = ∠F.

答案
$\because AB// CD,\therefore \angle DCF=\angle B.\because \angle B=\angle D,\therefore \angle DCF=\angle D.\therefore AD// BC.\therefore \angle DEF=\angle F$
3. 如图,EF⊥AB,CD⊥AB,垂足分别为E,D,点F在线段BC上,点G在线段AC上,∠EFB = ∠GDC. 求证:∠AGD = ∠ACB.

答案
$\because EF\bot AB,CD\bot AB,\therefore \angle BEF=\angle BDC = 90^{\circ }.\therefore EF// CD.\therefore \angle EFB=\angle BCD.\because \angle EFB=\angle GDC,\therefore \angle GDC=\angle BCD.\therefore DG// BC.\therefore \angle AGD=\angle ACB$
4.(教材P25习题7.3第3题变式)完成下面的证明推理过程,并在括号里填上依据.
如图,DE//BC,DF,BE分别平分∠ADE和∠ABC. 求证:∠FDE = ∠DEB.
证明:∵ DE//BC(已知),∴ ∠ADE = ________( ).
∵ DF,BE分别平分∠ADE和∠ABC(已知),
∴ ∠ADF = $\frac{1}{2}$∠ADE,∠ABE = $\frac{1}{2}$∠ABC( ).
∴ ∠ADF = ∠ABE.
∴ ________// ( ).
∴ ∠FDE = ________( ).
如图,DE//BC,DF,BE分别平分∠ADE和∠ABC. 求证:∠FDE = ∠DEB.
证明:∵ DE//BC(已知),∴ ∠ADE = ________( ).
∵ DF,BE分别平分∠ADE和∠ABC(已知),
∴ ∠ADF = $\frac{1}{2}$∠ADE,∠ABE = $\frac{1}{2}$∠ABC( ).
∴ ∠ADF = ∠ABE.
∴ ________// ( ).
∴ ∠FDE = ________( ).
答案
$\angle ABC$ 两直线平行,同位角相等 角的平分线的定义
DF BE 同位角相等,两直线平行 $\angle DEB$ 两直线平行,内错角相等
DF BE 同位角相等,两直线平行 $\angle DEB$ 两直线平行,内错角相等
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