登录
1. 如图，$AB=AC$，$AD=AE$，$BE=CD$，$∠2=110^{\circ}$，$∠BAE=60^{\circ}$，则下列结论错误的是 (
A.$△ABE≌△ACD$
B.$△ABD≌△ACE$
C.$∠ACE=30^{\circ}$
D.$∠1=70^{\circ}$
C
)A.$△ABE≌△ACD$
B.$△ABD≌△ACE$
C.$∠ACE=30^{\circ}$
D.$∠1=70^{\circ}$
答案
1.C
2. 如图，在四边形$ABCD$中，$AB=DC$，$AC=DB$，$AC$，$BD$相交于点$O$，则图中的全等三角形共有 (
A.2对
B.3对
C.4对
D.5对
B
)A.2对
B.3对
C.4对
D.5对
答案
2.B
解析
证明：在$\triangle ABC$和$\triangle DCB$中，
$\left\{\begin{array}{l} AB=DC\\ AC=DB\\ BC=CB\end{array}\right.$
$\therefore \triangle ABC≌\triangle DCB(SSS)$，
$\therefore \angle ABC=\angle DCB$，$\angle BAC=\angle CDB$，
在$\triangle ABD$和$\triangle DCA$中，
$\left\{\begin{array}{l} AB=DC\\ AD=DA\\ BD=CA\end{array}\right.$
$\therefore \triangle ABD≌\triangle DCA(SSS)$，
$\therefore \angle BAD=\angle CDA$，$\angle ABD=\angle DCA$，
在$\triangle AOB$和$\triangle DOC$中，
$\left\{\begin{array}{l} \angle BAO=\angle CDO\\ \angle AOB=\angle DOC\\ AB=DC\end{array}\right.$
$\therefore \triangle AOB≌\triangle DOC(AAS)$，
综上，全等三角形共有3对。
B
$\left\{\begin{array}{l} AB=DC\\ AC=DB\\ BC=CB\end{array}\right.$
$\therefore \triangle ABC≌\triangle DCB(SSS)$，
$\therefore \angle ABC=\angle DCB$，$\angle BAC=\angle CDB$，
在$\triangle ABD$和$\triangle DCA$中，
$\left\{\begin{array}{l} AB=DC\\ AD=DA\\ BD=CA\end{array}\right.$
$\therefore \triangle ABD≌\triangle DCA(SSS)$，
$\therefore \angle BAD=\angle CDA$，$\angle ABD=\angle DCA$，
在$\triangle AOB$和$\triangle DOC$中，
$\left\{\begin{array}{l} \angle BAO=\angle CDO\\ \angle AOB=\angle DOC\\ AB=DC\end{array}\right.$
$\therefore \triangle AOB≌\triangle DOC(AAS)$，
综上，全等三角形共有3对。
B
3. (新情境·现实生活)木工师傅在做完门框后为了防止变形，常用如图所示的方法钉上两根斜拉的木条，这样做的数学依据是
三角形具有稳定性
。答案
3.三角形具有稳定性
4. 如图，在四边形$ABCE$中，$AB=AC$，$AD=AE$，$BD=CE$，且$B$，$D$，$E$三点在同一条直线上。若$∠1=31^{\circ}$，$∠2=66^{\circ}$，则$∠3$的度数为
35°
。答案
4.35°
5. （2023·西藏）如图，$AB=DE$，$AC=DC$，$CE=CB$。求证：$∠1=∠2$。
答案
5.在△ABC和△DEC中，$\begin{cases}AB = DE, \\AC = DC, \\\ CB = CE,\end{cases}$
∴△ABC≌△DEC(SSS),
∴∠ACB = ∠DCE,
∴∠ACB - ∠ACE = ∠DCE - ∠ACE,
∴∠1 = ∠2
∴△ABC≌△DEC(SSS),
∴∠ACB = ∠DCE,
∴∠ACB - ∠ACE = ∠DCE - ∠ACE,
∴∠1 = ∠2
6. 如图，平面上有$△ACD$与$△BCE$，$AD$与$BE$相交于点$P$。若$AC=BC$，$AD=BE$，$CD=CE$，$∠ACE=55^{\circ}$，$∠BCD=155^{\circ}$，则$∠BPD$的度数为 (
A.$110^{\circ}$
B.$125^{\circ}$
C.$130^{\circ}$
D.$155^{\circ}$
C
)A.$110^{\circ}$
B.$125^{\circ}$
C.$130^{\circ}$
D.$155^{\circ}$
答案
6.C
解析
证明：在$\triangle ACD$和$\triangle BCE$中，
$\left\{\begin{array}{l} AC=BC\\ AD=BE\\ CD=CE\end{array}\right.$
$\therefore \triangle ACD≌\triangle BCE(SSS)$
$\therefore \angle ACD=\angle BCE$，$\angle A=\angle B$
$\because \angle ACD=\angle ACE+\angle ECD$，$\angle BCE=\angle BCD-\angle ECD$
$\therefore \angle ACE+\angle ECD=\angle BCD-\angle ECD$
$\because \angle ACE=55^{\circ}$，$\angle BCD=155^{\circ}$
$\therefore 55^{\circ}+\angle ECD=155^{\circ}-\angle ECD$
$\therefore 2\angle ECD=100^{\circ}$
$\therefore \angle ECD=50^{\circ}$
$\therefore \angle ACB=\angle ACE+\angle ECD+\angle DCB - \angle BCD$（此步表述有误，应为：$\angle ACB = \angle BCE - \angle ACE$，而$\angle BCE = \angle ACD = \angle ACE + \angle ECD = 55^{\circ} + 50^{\circ} = 105^{\circ}$，所以$\angle ACB = \angle BCE - \angle ACE = 105^{\circ} - 55^{\circ} = 50^{\circ}$，正确推导过程如下：）
$\because \angle ACD = \angle ACE + \angle ECD = 55^{\circ} + 50^{\circ} = 105^{\circ}$，且$\angle ACD = \angle BCE$
$\therefore \angle BCE = 105^{\circ}$
$\therefore \angle ACB = \angle BCE - \angle ACE = 105^{\circ} - 55^{\circ} = 50^{\circ}$
在$\triangle AQC$和$\triangle BQP$中，$\angle AQC = \angle BQP$，$\angle A = \angle B$
$\therefore \angle BPA = \angle ACB = 50^{\circ}$
$\therefore \angle BPD = 180^{\circ} - \angle BPA = 180^{\circ} - 50^{\circ} = 130^{\circ}$
答案：C
$\left\{\begin{array}{l} AC=BC\\ AD=BE\\ CD=CE\end{array}\right.$
$\therefore \triangle ACD≌\triangle BCE(SSS)$
$\therefore \angle ACD=\angle BCE$，$\angle A=\angle B$
$\because \angle ACD=\angle ACE+\angle ECD$，$\angle BCE=\angle BCD-\angle ECD$
$\therefore \angle ACE+\angle ECD=\angle BCD-\angle ECD$
$\because \angle ACE=55^{\circ}$，$\angle BCD=155^{\circ}$
$\therefore 55^{\circ}+\angle ECD=155^{\circ}-\angle ECD$
$\therefore 2\angle ECD=100^{\circ}$
$\therefore \angle ECD=50^{\circ}$
$\therefore \angle ACB=\angle ACE+\angle ECD+\angle DCB - \angle BCD$（此步表述有误，应为：$\angle ACB = \angle BCE - \angle ACE$，而$\angle BCE = \angle ACD = \angle ACE + \angle ECD = 55^{\circ} + 50^{\circ} = 105^{\circ}$，所以$\angle ACB = \angle BCE - \angle ACE = 105^{\circ} - 55^{\circ} = 50^{\circ}$，正确推导过程如下：）
$\because \angle ACD = \angle ACE + \angle ECD = 55^{\circ} + 50^{\circ} = 105^{\circ}$，且$\angle ACD = \angle BCE$
$\therefore \angle BCE = 105^{\circ}$
$\therefore \angle ACB = \angle BCE - \angle ACE = 105^{\circ} - 55^{\circ} = 50^{\circ}$
在$\triangle AQC$和$\triangle BQP$中，$\angle AQC = \angle BQP$，$\angle A = \angle B$
$\therefore \angle BPA = \angle ACB = 50^{\circ}$
$\therefore \angle BPD = 180^{\circ} - \angle BPA = 180^{\circ} - 50^{\circ} = 130^{\circ}$
答案：C
7. (易错题)如图，在$△ABC$中，$AB=AC$，$E$，$D$，$F$是$BC$的四等分点，$AE=AF$，则图中的全等三角形共有
4
对，分别是△ABE≌△ACF,△AED≌△AFD,△ABD≌△ACD,△ABF≌△ACE
。答案
7.4 △ABE≌△ACF,△AED≌△AFD,△ABD≌△ACD,△ABF≌△ACE [易错分析]解答本题时容易忽视△ABF≌△ACE,以致漏解.
解析
4；△ABE≌△ACF,△AED≌△AFD,△ABD≌△ACD,△ABF≌△ACE
