《2025年经纶学典课时作业八年级数学上册苏科版》

7. 如图,在$\triangle ABC$中,点D,E分别在AC,BC边上,$\triangle ADB \cong \triangle EDB$,$\triangle BDE \cong \triangle CDE$.下列结论:
①$BD是\angle ABE$的平分线;②$AB \perp AC$;③$\angle C= 30^\circ$;④线段$DE是\triangle BDC$的中线;⑤$AD+BD= AC$.其中正确的有( )

A.2个
B.3个
C.4个
D.5个

答案： D 解析:①$\because △ADB\cong △EDB,\therefore ∠ABD=∠EBD,$
∴BD 是$∠ABE$的平分线,故①正确;②$\because △BDE\cong △CDE,\therefore ∠DEB=∠DEC.\because ∠DEB+∠DEC=180^{\circ },\therefore ∠BED=90^{\circ }.\because △ADB\cong △EDB,\therefore ∠A=∠BED=90^{\circ },\therefore AB⊥AC$,故②正确;③$\because △ADB\cong △EDB,△BDE\cong △CDE,\therefore ∠ABD=∠EBD,∠EBD=∠C,\therefore ∠C=\frac {1}{3}(180^{\circ }-∠A)=\frac {1}{3}×90^{\circ }=30^{\circ }$,故③正确;④$\because △BDE\cong △CDE,\therefore BE=CE$,
∴线段 DE 是$△BDC$的中线,故④正确;⑤$\because △BDE\cong △CDE,\therefore BD=CD$,则$AD+BD=AD+CD=AC$,故⑤正确.故选 D.

8. 已知$\triangle ABC \cong \triangle DEF$,$AB= 5\ cm$,若$\triangle ABC的面积为10\ cm^2$,则$\triangle DEF的边DE$上的高为______$\ cm$.

答案： 4 解析:设$△ABC$的边 AB 上的高为 h cm,则$\frac {1}{2}AB\cdot h=10$,即$\frac {1}{2}×5h=10$,解得$h=4.\because △ABC\cong △DEF$,AB 与 DE 是对应边,
∴ DE 边上的高为 4 cm.

9. 如图,$\triangle ABC \cong \triangle ADE$,线段$BC$的延长线过点E,与线段$AD$交于点F,$\angle AED= 108^\circ$,$\angle CAD= 12^\circ$,$\angle B= 48^\circ$,则$\angle DEF$的度数为______.

答案： $36^{\circ }$ 解析:$\because △ABC\cong △ADE,\therefore ∠AED=∠ACB=108^{\circ }$.又$\because ∠B=48^{\circ },\therefore ∠CAB=180^{\circ }-∠B-∠ACB=180^{\circ }-48^{\circ }-108^{\circ }=24^{\circ }$.又$\because △ABC\cong △ADE,\therefore ∠EAD=∠CAB=24^{\circ }$.又$\because ∠EAB=∠EAD+∠CAD+∠CAB,∠CAD=12^{\circ },\therefore ∠EAB=24^{\circ }+12^{\circ }+24^{\circ }=60^{\circ },\therefore ∠AEB=180^{\circ }-∠EAB-∠B=180^{\circ }-60^{\circ }-48^{\circ }=72^{\circ },\therefore ∠DEF=∠AED-∠AEB=108^{\circ }-72^{\circ }=36^{\circ }.$

10. 如图,已知$\triangle ABD \cong \triangle CFD$,$AD \perp BC$于D.
(1)求证:$CE \perp AB$；
(2)已知$BC= 7,AD= 5$,求$AF$的长.

答案：
(1)$\because △ABD\cong △CFD,\therefore ∠BAD=∠FCD$.又$\because ∠AFE=∠CFD,\therefore ∠AEF=∠CDF=90^{\circ },\therefore CE⊥AB.$
(2)$\because △ABD\cong △CFD,\therefore BD=DF,AD=CD.\because BC=7,AD=DC=5,\therefore BD=BC-CD=2,\therefore AF=AD-DF=5-2=3.$

11. 如图,已知$\triangle ABC \cong \triangle DEF$,$CD平分\angle BCA$,若$\angle A= 30^\circ$,$\angle CGF= 88^\circ$,求$\angle E$的度数.

答案： $\because △ABC\cong △DEF,∠D=30^{\circ },\therefore ∠A=∠D=30^{\circ },∠E=∠B.\because ∠CGF=88^{\circ },\therefore ∠DCB=∠CGF-∠D=88^{\circ }-30^{\circ }=58^{\circ }.\because CD$平分$∠BCA,\therefore ∠ACB=2∠DCB=116^{\circ },\therefore ∠B=180^{\circ }-∠A-∠ACB=180^{\circ }-30^{\circ }-116^{\circ }=34^{\circ },\therefore ∠E=∠B=34^{\circ }.$

12. (几何直观·运算能力)如图,$\triangle ABE和\triangle ADC是\triangle ABC$分别沿着AB,AC边翻折$180^\circ$形成的,若$\angle 1:\angle 2:\angle 3= 28:5:3$,求$\angle \alpha$的度数.

答案： 根据题意设$∠1=28x,∠2=5x,∠3=3x$,则$28x+5x+3x=180^{\circ }$,解得$x=5^{\circ },\therefore ∠1=140^{\circ },∠2=25^{\circ },∠3=15^{\circ }$.由折叠的性质可知,$△ABE\cong △ADC\cong △ABC,\therefore ∠2=∠EBA=25^{\circ },∠3=∠ACD=15^{\circ },\therefore ∠EBC=50^{\circ },∠BCD=30^{\circ },\therefore ∠α=∠EBC+∠BCD=80^{\circ }.$