Question

如图，在Rt△ABC中，∠ACB=90°，∠B=30°，将△ABC绕点C按顺时针方向旋转n度后，得到△DEC，点D刚好落在AB边上．                                                                                               

（1）求n的值；                                                                                

（2）若F是DE的中点，判断四边形ACFD的形状，并说明理由．                

                                                           

Answer 1

【考点】旋转的性质；含30度角的直角三角形；直角三角形斜边上的中线；菱形的判定．               

【专题】几何图形问题．                                                                         

【分析】（1）利用旋转的性质得出AC=CD，进而得出△ADC是等边三角形，即可得出∠ACD的度数；               

（2）利用直角三角形的性质得出FC=DF，进而得出AD=AC=FC=DF，即可得出答案．              

【解答】解：（1）∵在Rt△ABC中，∠ACB=90°，∠B=30°，将△ABC绕点C按顺时针方向旋转n度后，得到△DEC，                                                                           

∴AC=DC，∠A=60°，                                                                       

∴△ADC是等边三角形，                                                                        

∴∠ACD=60°，                                                                                 

∴n的值是60；                                                                                 

                                                                                                          

（2）四边形ACFD是菱形；                                                                   

理由：∵∠DCE=∠ACB=90°，F是DE的中点，                                             

∴FC=DF=FE，                                                                                  

∵∠CDF=∠A=60°，                                                                          

∴△DFC是等边三角形，                                                                         

∴DF=DC=FC，                                                                                 

∵△ADC是等边三角形，                                                                        

∴AD=AC=DC，                                                                                 

∴AD=AC=FC=DF，                                                                           

∴四边形ACFD是菱形．                                                                         

【点评】此题主要考查了菱形的判定以及旋转的性质和直角三角形斜边上的中线等于斜边的一半等知识，得出△DFC是等边三角形是解题关键．