Question

如圖，在⊙O中，弦CD垂直于直徑AB于點F，OF=3，CD=8，M是OC的中點，AM的延長線交⊙O于點E，DE與BC交于點N，（1）求AB的長；（2）求證：BN=CN.

Answer 1

（1）10 （2）證明CN=，所以CN=BN

試題分析：（1）AB是⊙O直徑，AB⊥CD，CD=8
∴ CF=4                             
在Rt△OCF中，根據(jù)勾股定理，得
OC²=OF²+CF²                                  
=3²+4²
=25
∴OC=5                                      
∴AB=2OC=2×5=10                    
（2）連結(jié)AC，BD

∵弦CD垂直于直徑AB，
∴BC=BD．
∴∠BCD=∠BDC   
∵OA=OC，                                                          
∴∠OCA=∠OAC．                                                                    
∵∠BDC=∠OAC，                                                                    
∴∠BCD=∠OCA．                                                                        
∴△BCD∽△OCA                                                                          
∴                                                                       
在△CDN和△CAM中，                                                                    
∵∠DCN=∠ACM，∠CDN=∠CAM，                                                       
∴△CDN∽△CAM                                                                          
∴
∴  
∴
即BN=CN．  
點評：本題考查勾股定理，相似三角形，解答本題的關(guān)鍵是掌握勾股定理的內(nèi)容，熟悉相似三角形的判定方法