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