Question

已知：如图，在△ABC中，∠ABC＝90°，以AB上的点O为圆心，OB的长为半径的圆与AB交于点E，与AC切于点D．

1.求证：BC＝CD；

2.求证：∠ADE＝∠ABD；

3.设AD＝2，AE＝1，求⊙O直径的长．

 

Answer 1

 

1.∵∠ABC＝90°，

∴OB⊥BC．······························································· 1分

∵OB是⊙O的半径，

∴CB为⊙O的切线．·················································· 2分

又∵CD切⊙O于点D，

∴BC＝CD；  

2.∵BE是⊙O的直径，

∴∠BDE＝90°．

∴∠ADE＋∠CDB ＝90°．······································ 4分

又∵∠ABC＝90°，

∴∠ABD＋∠CBD＝90°．··········································································· 5分

由（1）得BC＝CD，∴∠CDB ＝∠CBD．

∴∠ADE＝∠ABD；         6分

3.由（2）得，∠ADE＝∠ABD，∠A＝∠A．

∴△ADE∽△ABD．··················································································· 7分

∴＝．·························································································· 8分

∴＝，∴BE＝3，············································································ 9分

∴所求⊙O的直径长为3．        10分

 解析:略