题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020833-60765.jpg)
求证:AM=AN.
答案
解析
证明:∵△AEB由△ADC旋转而得,
∴△AEB≌△ADC,
∴∠EAB=∠CAD,∠EBA=∠C,
∵AB=AC,AD⊥BC,
∴∠BAD=∠CAD,∠ABC=∠C,
∴∠EAB=∠DAB,
∠EBA=∠DBA,
∵∠EBM=∠DBN,
∴∠MBA=∠NBA,
又∵AB=AB,
∴△AMB≌△ANB(ASA),
∴AM=AN.
核心考点
试题【如图,在△ABC中,AB=AC,AD⊥BC于点D,将△ADC绕点A顺时针旋转,使AC与AB重合,点D落在点E处,AE的延长线交CB的延长线于点M,EB的延长线交】;主要考察你对相似图形性质等知识点的理解。[详细]
举一反三
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020829-39791.jpg)
(1)当∠BQD=30°时,求AP的长;
(2)在运动过程中线段ED的长是否发生变化?如果不变,求出线段ED的长;如果变化请说明理由.
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020823-30849.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020823-40001.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020823-52172.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020824-89124.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020824-12495.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020824-36677.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020824-58559.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103020825-67770.jpg)