Question

如圖，在平行四邊形ABCD中，對角線AC、BD交于點O．M為AD中點，連接CM交BD于點N，且ON=1．                                                 

（1）求BD的長；                                                                             

（2）若△DCN的面積為2，求四邊形ABNM的面積．                                   

                                                               

Answer 1

【考點】相似三角形的判定與性質(zhì)；平行四邊形的性質(zhì)．                               

【專題】幾何綜合題．                                                                       

【分析】（1）由四邊形ABCD為平行四邊形，得到對邊平行且相等，且對角線互相平分，根據(jù)兩直線平行內(nèi)錯角相等得到兩對角相等，進(jìn)而確定出三角形MND與三角形CNB相似，由相似得比例，得到DN：BN=1：2，設(shè)OB=OD=x，表示出BN與DN，求出x的值，即可確定出BD的長；                                            

（2）由相似三角形相似比為1：2，得到CN=2MN，BN=2DN．已知△DCN的面積，則由線段之比，得到△MND與△CNB的面積，從而得到S_△ABD=S_△BCD=S_△BCN+S_△CND，最后由S_{四邊形ABNM}=S_△ABD﹣S_△MND求解．                  

【解答】解：（1）∵平行四邊形ABCD，                                               

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

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

∴△MND∽△CNB，                                                                         

∴=，                                                                                        

∵M(jìn)為AD中點，                                                                              

∴MD=AD=BC，即=，                                                                 

∴=，即BN=2DN，                                                                          

設(shè)OB=OD=x，則有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，且相似比為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_{四邊形ABNM}=S_△ABD﹣S_△MND=6﹣1=5．                                                 

【點評】此題考查了相似三角形的判定與性質(zhì)，熟練掌握相似三角形的判定與性質(zhì)是解本題的關(guān)鍵．