题目内容
如图,AB是⊙O的直径,点D在AB的延长线上,点C在⊙O上, CA=CD,∠CDA=30°.
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(1)试判断直线CD与⊙O的位置关系,并说明理由;
(2)若⊙O的半径为4,求点A到CD所在直线的距离
(1)证明略
(2)6
(1)
△ACD是等腰三角形,∠D=30°.
∠CAD=∠CDA=30°.
连接OC,
AO=CO,
△AOC是等腰三角形.
∠CAO=∠ACO=30°,
∠COD=60°.
在△COD中,又
∠CDO=30°,
∠DCO=90°
CD是⊙O的切线,即直线CD与⊙O相切.
(2)过点A 作AE⊥CD,垂足为E.
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在Rt△COD中,
∠CDO=30°,
OD=2OC=8. AD="AO+OD=12"
在Rt△ADE中,
∠EDA=30°,
点A到CD边的距离为: AE="AD/2=6"
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连接OC,
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在△COD中,又
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(2)过点A 作AE⊥CD,垂足为E.
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在Rt△COD中,
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在Rt△ADE中,
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