《2025年能力培养与测试八年级数学上册人教版》

9. 如图14.2 - 9,在△ABC中,D是边BC的中点,连接AD并延长到点E,使DE = AD,连接CE.
(1)求证△ABD≌△ECD；
(2)若△ABD的面积为5,求△ACE的面积.

答案： （1）证明：$\because D$是边$BC$的中点，$\therefore BD=CD$．在$\triangle ABD$和$\triangle ECD$中，$\left\{\begin{array}{l} BD=CD,\\ \angle ADB=\angle EDC,\\ AD=ED,\end{array}\right. $$\therefore \triangle ABD≌\triangle ECD$（SAS）．（2）解：$\because D$是边$BC$的中点，$\therefore S_{\triangle ABD}=S_{\triangle ACD}$．$\because \triangle ABD≌\triangle ECD$，$\therefore S_{\triangle ABD}=S_{\triangle ECD}$．$\therefore S_{\triangle ACD}=S_{\triangle ECD}=S_{\triangle ABD}=5$．$\therefore S_{\triangle ACE}=S_{\triangle ACD}+S_{\triangle ECD}=5+5=10$，即$\triangle ACE$的面积为10．

10. 如图14.2 - 10,在△ABC中,AB = 7,AC = 3,AD是△ABC的中线,求AD的长的取值范围.

答案：
解：如图，延长$AD$到点$E$，使$DE=DA$，连接$BE$． $\because AD$是$\triangle ABC$的中线，$\therefore BD=CD$．在$\triangle BED$和$\triangle CAD$中，$\left\{\begin{array}{l} BD=CD,\\ \angle EDB=\angle ADC,\\ DE=DA,\end{array}\right. $$\therefore \triangle BED≌\triangle CAD$（SAS）．$\therefore EB=AC=3$．$\because ED=AD$，$\therefore AE=2AD$．在$\triangle ABE$中，$AB - EB\lt AE\lt AB + EB$，$\therefore 7 - 3\lt 2AD\lt 7 + 3$．$\therefore 2\lt AD\lt 5$．

11. 如图14.2 - 11,在△ABC中,∠BAC = ∠B = 60°,AB = AC,D,E分别是边BC,AB所在直线上的动点,且BD = AE,直线AD与EC交于点F.
(1)当点D,E分别在边BC,AB上运动时,∠DFC的度数是否发生变化？若不变,求出其度数；若变化,写出其变化规律.
(2)当点D,E分别运动到CB,BA的延长线上时,(1)中的结论是否改变？请说明理由.

答案：
（1）$\angle DFC$的度数不发生变化，且$\angle DFC=60^{\circ }$．在$\triangle ABD$和$\triangle CAE$中，$\left\{\begin{array}{l} AB=CA,\\ \angle B=\angle CAE,\\ BD=AE,\end{array}\right. $$\therefore \triangle ABD≌\triangle CAE$（SAS）．$\therefore \angle BAD=\angle ACE$．$\therefore \angle DFC=\angle DAC+\angle ACE=\angle DAC+\angle BAD=\angle BAC=60^{\circ }$．（2）（1）中的结论不改变．理由如下：当点$D$，$E$分别运动到$CB$，$BA$的延长线上时，如图． $\because \angle ABD+\angle ABC=180^{\circ }$，$\angle CAE+\angle BAC=180^{\circ }$，$\angle ABC=\angle BAC=60^{\circ }$，$\therefore \angle ABD=\angle CAE=120^{\circ }$．在$\triangle ABD$和$\triangle CAE$中，$\left\{\begin{array}{l} AB=CA,\\ \angle ABD=\angle CAE,\\ BD=AE,\end{array}\right. $$\therefore \triangle ABD≌\triangle CAE$（SAS）．$\therefore \angle D=\angle E$．$\because \angle EAF=\angle BAD$，$\therefore \angle DFC=\angle EAF+\angle E=\angle BAD+\angle D=\angle ABC=60^{\circ }$．$\therefore$当点$D$，$E$分别运动到$CB$，$BA$的延长线上时，（1）中的结论不改变．