《2025年暑假衔接八年级数学浙教版延边人民出版社》

17. 已知DH是BC边的垂直平分线,AD是$△ABC$的外角$∠BAG$的平分线,两线交于点D,$DF⊥AB$,垂足为F,$DE⊥CG$,垂足为E.
(1)求证:$BF=CE$.
(2)若$AC=4cm,AB=8cm$,求AE的长.


(1)证明:连结 BD,CD. ∵DH 是 BC 边的垂直平分线,$\therefore DB=DC$; ∵DA 平分$∠BAG,DE⊥CG,DF⊥AB,\therefore DE=DF,∠DEC=∠DFB=90^{\circ },\therefore Rt△DEC≌Rt△DFB(HL),\therefore BF=CE$.
(2)$\because AD=AD,DE=DF,∠DEA=∠DFA=90^{\circ },\therefore Rt△DEA≌Rt△DFA(HL),\therefore AE=AF,\therefore AE=AF=AB - BF=AB - CE=AB - AC - AE,\therefore 2AE=AB - AC=8 - 4=4(cm),\therefore AE=$2cm

答案：
(1)证明:连结 BD,CD.
∵DH 是 BC 边的垂直平分线,$\therefore DB=DC$;
∵DA 平分$∠BAG,DE⊥CG,DF⊥AB,\therefore DE=DF,∠DEC=∠DFB=90^{\circ },\therefore Rt△DEC≌Rt△DFB(HL),\therefore BF=CE$.
(2)$\because AD=AD,DE=DF,∠DEA=∠DFA=90^{\circ },\therefore Rt△DEA≌Rt△DFA(HL),\therefore AE=AF,\therefore AE=AF=AB - BF=AB - CE=AB - AC - AE,\therefore 2AE=AB - AC=8 - 4=4(cm),\therefore AE=2cm$

18. 折纸是我国一项古老的传统民间艺术,这项具有中国特色的传统文化在几何中可以得到新的解读. 请根据题意,探索不同情境中$∠1+∠2$(或$∠1-∠2$)与$∠A$的数量关系.
(1)如图1,已知在$△ABC$中,$∠A=60^{\circ }$,若沿图中虚线DE截去$∠A$,则$∠1+∠2=$______.
(2)如图2,翻折后,点A落在点$A'$处,若$∠1+∠2=110^{\circ }$,求$∠B+∠C$的度数.
(3)如图3,将$△ABC$沿DE折叠,使点A落在点$A'$处,若$∠1=80^{\circ },∠2=28^{\circ }$,则$∠A$的度数为______.

答案：

(1)$\because ∠A=60^{\circ },\therefore ∠ADE+∠AED=180^{\circ }-60^{\circ }=120^{\circ },\therefore ∠1+∠2=360^{\circ }-∠ADE-∠AED=240^{\circ }$,故答案为:$240^{\circ }$.
(2)连结$AA'$,如图1所示,$\because ∠1=∠DAA'+∠DA'A,∠2=∠EAA'+∠EA'A,\therefore ∠1+∠2=∠DAA'+∠DA'A+∠EAA'+∠EA'A=∠EAD+∠EA'D$. $\because ∠EAD=∠EA'D,\therefore ∠1+∠2=2∠EAD=110^{\circ },\therefore ∠EAD=55^{\circ },\therefore ∠B+∠C=180^{\circ }-55^{\circ }=125^{\circ }$.
(3)如图2,设 AB 与$DA'$交于点 F,$\because ∠1=∠DFA+∠A,∠DFA=∠2+∠A'$,由折叠可得,$\angle A=\angle {A}^{\prime },\therefore \angle 1=\angle A+\angle {A}^{\prime }+\angle 2=2\angle A+∠2$. 又$\because ∠1=80^{\circ },∠2=28^{\circ },\therefore 80^{\circ }=2∠A+28^{\circ },\therefore ∠A=26^{\circ }$. 故答案为：$26^{\circ }$.