Question

已知：如圖，在△ABC中，∠ABC＝90°，以AB上的點(diǎn)O為圓心，OB的長為半徑的圓與AB交于點(diǎn)E，與AC切于點(diǎn)D．

1.求證：BC＝CD；

2.求證：∠ADE＝∠ABD；

3.設(shè)AD＝2，AE＝1，求⊙O直徑的長．

 

Answer 1

 

1.∵∠ABC＝90°，

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

∵OB是⊙O的半徑，

∴CB為⊙O的切線．·················································· 2分

又∵CD切⊙O于點(diǎn)D，

∴BC＝CD；  

2.∵BE是⊙O的直徑，

∴∠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的直徑長為3．        10分

 解析:略