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

p;【答案】
【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分解析:
p;【解析】略