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．

【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

p;【答案】
【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(diǎn)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分解析:
p;【解析】略