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．

小题1:求证：BC＝CD；
小题2:求证：∠ADE＝∠ABD；
小题3:设AD＝2，AE＝1，求⊙O直径的长．

Answer 1

 
小题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；      
小题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
小题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

 略