Question

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

小題1:求證：BC＝CD；
小題2:求證：∠ADE＝∠ABD；
小題3:設(shè)AD＝2，AE＝1，求⊙O直徑的長．

Answer 1

 
小題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D，
∴BC＝CD；      
小題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
小題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

 略