Question

如图，在平行四边形ABCD中，对角线AC、BD交于点O．M为AD中点，连接CM交BD于点N，且ON=1．                                                 

（1）求BD的长；                                                                             

（2）若△DCN的面积为2，求四边形ABNM的面积．                                   

                                                               

Answer 1

【考点】相似三角形的判定与性质；平行四边形的性质．                               

【专题】几何综合题．                                                                       

【分析】（1）由四边形ABCD为平行四边形，得到对边平行且相等，且对角线互相平分，根据两直线平行内错角相等得到两对角相等，进而确定出三角形MND与三角形CNB相似，由相似得比例，得到DN：BN=1：2，设OB=OD=x，表示出BN与DN，求出x的值，即可确定出BD的长；                                            

（2）由相似三角形相似比为1：2，得到CN=2MN，BN=2DN．已知△DCN的面积，则由线段之比，得到△MND与△CNB的面积，从而得到S_△ABD=S_△BCD=S_△BCN+S_△CND，最后由S_{四边形ABNM}=S_△ABD﹣S_△MND求解．                  

【解答】解：（1）∵平行四边形ABCD，                                               

∴AD∥BC，AD=BC，OB=OD，                                                              

∴∠DMN=∠BCN，∠MDN=∠NBC，                                                     

∴△MND∽△CNB，                                                                         

∴=，                                                                                        

∵M为AD中点，                                                                              

∴MD=AD=BC，即=，                                                                 

∴=，即BN=2DN，                                                                          

设OB=OD=x，则有BD=2x，BN=OB+ON=x+1，DN=x﹣1，                           

∴x+1=2（x﹣1），                                                                           

解得：x=3，                                                                                      

∴BD=2x=6；                                                                                     

                                                                                                          

（2）∵△MND∽△CNB，且相似比为1：2，                                         

∴MN：CN=DN：BN=1：2，                                                                  

∴S_△MND=S_△CND=1，S_△BNC=2S_△CND=4．                                             

∴S_△ABD=S_△BCD=S_△BCN+S_△CND=4+2=6                                                   

∴S_{四边形ABNM}=S_△ABD﹣S_△MND=6﹣1=5．                                                 

【点评】此题考查了相似三角形的判定与性质，熟练掌握相似三角形的判定与性质是解本题的关键．