《2025年名校课堂八年级数学上册人教版云南专版》

7.如图，∠ABC=∠DEF，AB=DE，试说明：△ABC≌△DEF.
(1)若以“SAS”为依据，还需添加的条件为$BC = EF$或$BE = CF$；
(2)若以“ASA”为依据，还需添加的条件为$\angle A = \angle D$；
(3)若以“AAS”为依据，还需添加的条件为$\angle ACB = \angle DFE$.

答案： 7.
(1)$BC = EF$或$BE = CF$
(2)$\angle A = \angle D$
(3)$\angle ACB = \angle DFE$

8.(昆明期末)如图，某同学把一块三角形的玻璃打碎成了3块，现在要到玻璃店去配一块大小完全一样的玻璃，那么最省事的方法是(C)

A.带①去
B.带②去
C.带③去
D.①②③都带去

答案： 8.C

9.如图，在△ACD中，∠CAD=90°，AC=6，AD=8，AB//CD，E是CD上一点，BE交AD于点F.若AB=DE，则图中阴影部分的面积为 (A)

A.24
B.30
C.42
D.48

答案： 9.A

10.如图，在Rt△ABC和Rt△ADE中，∠C=∠E=90°，∠CAD=∠EAB，AC=AE，AB，DE相交于点F，AD，BC相交于点G.
(1)求证：△ABC≌△ADE；
(2)若AB=11，AG=6，求DG的长.

答案： 10.解：
(1)证明：$\because \angle CAD = \angle EAB$，$\therefore \angle CAD + \angle BAD = \angle EAB + \angle DAB$，即$\angle CAB = \angle EAD$. $\because AC = AE$，$\angle C = \angle E = 90^{\circ}$，$\therefore \triangle ABC \cong \triangle ADE$(ASA).
(2)$\because \triangle ABC \cong \triangle ADE$，$\therefore AB = AD$. $\because AB = 11$，$\therefore AD =$
$11$.又$\because AG = 6$，$\therefore DG = 11 - 6 = 5$.

11.(曲靖期末)如图1，已知点A是线段DE上一点，∠BAC=90°，AB=AC，BD⊥DE，CE⊥DE.
(1)求证：DE=BD+CE；
(2)如果是如图2这个图形，我们能得到什么结论？并证明.
　　　　

答案： 11.解：
(1)证明：$\because BD \perp DE$，$CE \perp DE$，$\therefore \angle D = \angle E = 90^{\circ}$. $\therefore \angle DBA + \angle DAB = 90^{\circ}$. $\because \angle BAC = 90^{\circ}$，$\therefore \angle DAB + \angle CAE = 90^{\circ}$. $\therefore \angle DBA = \angle CAE$.又$\because AB = AC$，$\therefore \triangle ADB \cong \triangle CEA$(AAS). $\therefore BD = AE$，$CE = AD$. $\therefore DE = AD + AE = CE + BD$.
(2)结论：$BD = CE + DE$.证明：$\because BD \perp DE$，$CE \perp DE$，$\therefore \angle ADB = \angle AEC = 90^{\circ}$. $\therefore \angle ABD + \angle BAD = 90^{\circ}$. $\because \angle BAC = 90^{\circ}$，$\therefore \angle BAD + \angle EAC = 90^{\circ}$. $\therefore \angle ABD = \angle EAC$.又$\because AB = AC$，$\therefore \triangle ADB \cong \triangle CEA$(AAS). $\therefore BD = AE$，$CE = AD$.又$\because AE = AD +$
$DE$，$\therefore BD = CE + DE$.