题目内容
如图,梯形ABCD内接于⊙O,AD∥BC,过点C作⊙O的切线,交BD的延长线于点P,交AD的延长线于点E.
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(1)求证:AB2=DE·BC;
(2)若BD=9,AB=6,BC=9,求切线PC的长.
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(1)求证:AB2=DE·BC;
(2)若BD=9,AB=6,BC=9,求切线PC的长.
(1)见解析(2)
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(1)∵AD∥BC,∴
.∴AB=CD,∠EDC=∠BCD.
又PC与⊙O相切,∴∠ECD=∠DBC.∴△CDE∽△BCD.∴
.
∴CD2=DE·BC,即AB2=DE·BC.
(2)由(1)知,DE=
=4,∵AD∥BC,∴△PDE∽△PBC,∴
又∵PB-PD=9,∴PD=
,PB=
.∴PC2=PD·PB=
·
=
.∴PC=
.
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又PC与⊙O相切,∴∠ECD=∠DBC.∴△CDE∽△BCD.∴

∴CD2=DE·BC,即AB2=DE·BC.
(2)由(1)知,DE=
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