题目内容
(2006•湖北)如图,已知CA、CB都经过点C,AC是⊙B的切线,⊙B交AB于点D,连接CD并延长交OA于点E,连接AF.(1)求证:AE⊥AB;
(2)求证:DE•DC=2AD•DB;
(3)如果AE=3,BD=4,求DC的长.
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【答案】分析:(1)根据切线的性质知:∠ACB=90°,由AC=AE,可得∠AEC=∠ACE,由BC=BD,可知∠BCD=∠BDC,再根据∠BDC=∠ADE,可得AE⊥AB;
(2)根据△ADE∽△FDB可得出DE•DC=2AD•DB;
(3)在Rt△ABC中,根据勾股定理可将AD的长求出,代入第二个小题的结论,可得出DC的长.
解答:
(1)证明:∵AC是⊙B的切线,
∴∠ACB=∠ACD+∠BCD=90°.
∵BC=BD,
∴∠BCD=∠BDC.
∴∠ACD+∠BDC=90°.
∵AC=AE,
∴∠ACD=∠AED.
∵∠ADE=∠BCD,
∴∠AED+∠ADE=90°.
∴∠EAD=90°.
即AE⊥AB.
(2)证明:过点B作BF⊥CD于点F,
∵∠ADE=∠BDF,∠EAD=∠BFD,
∴△ADE∽△FDB.
∴
=
.
即DE•FD=AD•DB.
∵DC=2FD,
∴DE•DC=2AD•DB.
(3)解:∵AE=3,BD=4,
在Rt△ABC中,
(AD+BD)2=AC2+BC2.
即(AD+4)2=32+42解得AD=1,
∴DE=
=
=
.
∵DE•DC=2AD•DB,
即
×DC=2×1×4,
∴DC=
.
点评:本题主要考查切线的性质及相似三角形的判定和应用.
(2)根据△ADE∽△FDB可得出DE•DC=2AD•DB;
(3)在Rt△ABC中,根据勾股定理可将AD的长求出,代入第二个小题的结论,可得出DC的长.
解答:
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∴∠ACB=∠ACD+∠BCD=90°.
∵BC=BD,
∴∠BCD=∠BDC.
∴∠ACD+∠BDC=90°.
∵AC=AE,
∴∠ACD=∠AED.
∵∠ADE=∠BCD,
∴∠AED+∠ADE=90°.
∴∠EAD=90°.
即AE⊥AB.
(2)证明:过点B作BF⊥CD于点F,
∵∠ADE=∠BDF,∠EAD=∠BFD,
∴△ADE∽△FDB.
∴
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即DE•FD=AD•DB.
∵DC=2FD,
∴DE•DC=2AD•DB.
(3)解:∵AE=3,BD=4,
在Rt△ABC中,
(AD+BD)2=AC2+BC2.
即(AD+4)2=32+42解得AD=1,
∴DE=
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∵DE•DC=2AD•DB,
即
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∴DC=
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点评:本题主要考查切线的性质及相似三角形的判定和应用.
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