Question

如圖，在Rt△ABC中，∠ACB=90°，∠B=30°，將△ABC繞點(diǎn)C按順時(shí)針?lè)较蛐�轉(zhuǎn)n度后，得到△DEC，點(diǎn)D剛好落在AB邊上．                                                                                               

（1）求n的值；                                                                                

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

                                                           

Answer 1

【考點(diǎn)】旋轉(zhuǎn)的性質(zhì)；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(wèn)題．                                                                         

【分析】（1）利用旋轉(zhuǎn)的性質(zhì)得出AC=CD，進(jìn)而得出△ADC是等邊三角形，即可得出∠ACD的度數(shù)；               

（2）利用直角三角形的性質(zhì)得出FC=DF，進(jìn)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(diǎn)C按順時(shí)針?lè)较蛐�轉(zhuǎn)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(xiàn)是DE的中點(diǎn)，                                             

∴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(diǎn)評(píng)】此題主要考查了菱形的判定以及旋轉(zhuǎn)的性質(zhì)和直角三角形斜邊上的中線等于斜邊的一半等知識(shí)，得出△DFC是等邊三角形是解題關(guān)鍵．