如图，在同一平面内，将△ABC绕点A旋转到△AED的位置，若AE⊥BC，∠ADC=65°，则∠ABC的度数为（...

问题详情：

如图，在同一平面内，将△ABC绕点A旋转到△AED的位置，若AE⊥BC，∠ADC=65°，则∠ABC的度数为（　　）                                                                                     

A．30°                              B．40°                               C．50°                               D．60°

【回答】

B【考点】旋转的*质．                                                                                     

【分析】先根据旋转的*质得AD=AC，∠BAE=∠CAD，再根据等腰三角形的*质和三角形内角和计算出∠CAD=50°，则∠BAE=50°，然后利用互余计算∠ABC的度数．                                                        

【解答】解：∵△ABC绕点A旋转到△AED的位置，                                            

∴AD=AC，∠BAE=∠CAD，                                                                                 

∵AD=AC，                                                                                                        

∴∠ACD=∠ADC=65°，                                                                                     

∴∠CAD=180°﹣65°﹣65°=50°，                                                                           

∴∠BAE=50°，                                                                                                   

∵AE⊥BC，                                                                                                       

∴∠ABC=90°﹣∠BAE=40°．                                                                                 

知识点：图形的旋转

题型：选择题