Question

如圖，在Rt△ABC中，∠ACB=90°，∠B=30°，將△ABC繞點C按順時針方向旋轉n度后，得到△DEC，點D剛好落在AB邊上．                                                                                               

（1）求n的值；                                                                                

（2）若F是DE的中點，判斷四邊形ACFD的形狀，并說明理由．                

                                                           

Answer 1

【考點】旋轉的性質；含30度角的直角三角形；直角三角形斜邊上的中線；菱形的判定．               

【專題】幾何圖形問題．                                                                         

【分析】（1）利用旋轉的性質得出AC=CD，進而得出△ADC是等邊三角形，即可得出∠ACD的度數；               

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

【解答】解：（1）∵在Rt△ABC中，∠ACB=90°，∠B=30°，將△ABC繞點C按順時針方向旋轉n度后，得到△DEC，                                                                           

∴AC=DC，∠A=60°，                                                                       

∴△ADC是等邊三角形，                                                                        

∴∠ACD=60°，                                                                                 

∴n的值是60；                                                                                 

                                                                                                          

（2）四邊形ACFD是菱形；                                                                   

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

∴FC=DF=FE，                                                                                  

∵∠CDF=∠A=60°，                                                                          

∴△DFC是等邊三角形，                                                                         

∴DF=DC=FC，                                                                                 

∵△ADC是等邊三角形，                                                                        

∴AD=AC=DC，                                                                                 

∴AD=AC=FC=DF，                                                                           

∴四邊形ACFD是菱形．                                                                         

【點評】此題主要考查了菱形的判定以及旋轉的性質和直角三角形斜邊上的中線等于斜邊的一半等知識，得出△DFC是等邊三角形是解題關鍵．