3. 请将下面的说理过程补充完整.
如图,$AB⊥ AC$,$DE⊥ AC$,$∠ B=∠ D$.请说明$AD// BC$的理由.

理由: $\because AB⊥ AC$,$DE⊥ AC$,(已知)
$\therefore$
$\therefore$
$\therefore ∠ B=∠ DEC$(
又$\because ∠ B=∠ D$(已知),
$\therefore ∠ D=$
$\therefore AD// BC$(
如图,$AB⊥ AC$,$DE⊥ AC$,$∠ B=∠ D$.请说明$AD// BC$的理由.
理由: $\because AB⊥ AC$,$DE⊥ AC$,(已知)
$\therefore$
$∠ BAC$
$=$$∠ EFC$
$=90°$(垂直的定义).$\therefore$
$AB$
$//$$DE$
(同位角相等,两直线平行
).$\therefore ∠ B=∠ DEC$(
两直线平行,同位角相等
).又$\because ∠ B=∠ D$(已知),
$\therefore ∠ D=$
$∠ DEC$
(等量代换).$\therefore AD// BC$(
内错角相等,两直线平行
).答案
$∠ BAC$;$∠ EFC$;$AB$;$DE$;同位角相等,
两直线平行;两直线平行,同位角相等;
$∠ DEC$;内错角相等,两直线平行
两直线平行;两直线平行,同位角相等;
$∠ DEC$;内错角相等,两直线平行
4. 请将下面的解题过程补充完整.
如图,$EF// AD$,$∠ BAC=70°$,$∠ 1=∠ 2$.求$∠ AGD$的度数.

解: $\because EF// AD$,
$\therefore ∠ 2=$
$\because ∠ 1=∠ 2$,
$\therefore ∠ 1=∠ 3$.
$\therefore$
$\therefore ∠ BAC+$
$\because ∠ BAC=70°$,
$\therefore ∠ AGD=$
如图,$EF// AD$,$∠ BAC=70°$,$∠ 1=∠ 2$.求$∠ AGD$的度数.
解: $\because EF// AD$,
$\therefore ∠ 2=$
$∠ 3$
(两直线平行,同位角相等
).$\because ∠ 1=∠ 2$,
$\therefore ∠ 1=∠ 3$.
$\therefore$
$DG$
$//$$AB$
(内错角相等,两直线平行
).$\therefore ∠ BAC+$
$∠ AGD$
$=180°$(两直线平行,同旁内角互补
).$\because ∠ BAC=70°$,
$\therefore ∠ AGD=$
$110°$
.答案
$∠ 3$;两直线平行,同位角相等;$DG$;
$AB$;内错角相等,两直线平行;$∠ AGD$;
两直线平行,同旁内角互补;$110°$
$AB$;内错角相等,两直线平行;$∠ AGD$;
两直线平行,同旁内角互补;$110°$
5. 如图,$AB// CD$,AE平分$∠ BAD$,CD与AE相交于点F,$∠ CFE=∠ E$.请说明$AD// BC$的理由.

答案
理由:$\because AE$平分$∠ BAD$,
$\therefore ∠ 1=∠ 2$.
$\because AB// CD$,$∠ CFE=∠ E$,
$\therefore ∠ 1=∠ CFE=∠ E$.
$\therefore ∠ 2=∠ E$.
$\therefore AD// BC$.
$\therefore ∠ 1=∠ 2$.
$\because AB// CD$,$∠ CFE=∠ E$,
$\therefore ∠ 1=∠ CFE=∠ E$.
$\therefore ∠ 2=∠ E$.
$\therefore AD// BC$.
6. 如图,$DE⊥ AC$,$∠ AGF=∠ ABC$,$∠ 1+∠ 2=180°$.请判断BF与AC的位置关系,并说明理由.

答案
解:$BF⊥ AC$.
理由:$\because ∠ AGF=∠ ABC$,
$\therefore BC// GF$.
$\therefore ∠ 1=∠ 3$.
又$\because ∠ 1+∠ 2=180°$,
$\therefore ∠ 2+∠ 3=180°$.
$\therefore BF// DE$.
$\because DE⊥ AC$,
$\therefore BF⊥ AC$.
理由:$\because ∠ AGF=∠ ABC$,
$\therefore BC// GF$.
$\therefore ∠ 1=∠ 3$.
又$\because ∠ 1+∠ 2=180°$,
$\therefore ∠ 2+∠ 3=180°$.
$\therefore BF// DE$.
$\because DE⊥ AC$,
$\therefore BF⊥ AC$.
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