已知:如图28-Y-14,⊙O的直径AB垂直于弦CD,过点C的切线与直径AB

已知:如图28Y14O的直径AB垂直于弦CD,过点C的切线与直径AB的延长线相交于点P,连接PD.

(1)求证:PDO的切线;

(2)求证:PD2PB·PA

(3)PD4tanCDB,求直径AB的长.

             

答案

解:(1)证明:连接ODOC.

PCO的切线,

OCPC

∴∠OCP90°.

直径ABCD

OPCD垂直平分线上的点,

ODOCPDPC.

OPOP

∴△ODP≌△OCP

∴∠ODPOCP90°.

ODO的半径,

PDO的切线.

(2)证明:∵∠ODP90°

∴∠PDBODB90°.

AB是直径,∴∠ADB90°

∴∠ADOODB90°

∴∠PDBADOA.

∵∠DPBAPD

∴△DPB∽△APD

PDPAPBPD

PD2PB·PA.

(3)∵∠AABD90°CDBABD

∴∠ACDB.

tanCDB

tanA

AD2BD.

∵△DPB∽△APD

PDPAPBPDBDDA1     2.

PD4

PA8PB2

AB6.

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