Question

如图2-5-19,已知PA为⊙O的切线,PO交⊙O于点B,BC⊥PA于点C,交⊙O于点D,

图2-5-19

(1)求证:AB²=PB·BD.

(2)若PA =15,PB =5,求BD的长.

Answer 1

思路分析:(1)只需证△PBA∽△ABD.?

(2)在(1)的基础上,只需求AB,因此寻找AB与BE的关系式,这可以通过相似三角形和勾股定理达到目的.

(1)证明:连结AD,延长PO交⊙O于E,连结AE.?

∵BC⊥PA,∴∠P +∠PBC =90°.?

∵BE为直径,?

∴∠BAE =90°,∠BAD +DAE =90°.?

∵∠DAE =∠DBE =∠PBC,∴∠P =∠BAD.?

又∵∠PAB =∠ADB,∴△PBA∽△ABD.?

∴=,即AB² =PB·BD.

(2)解:∵PA为切线,∴PA²=PB·PE.?

又PA =15,PB =5,∴PE =45.?

∴BE =40.?

∵△PBA∽△PAE,∴= ==.?

设AB =x,则AE =3x.

又AB²+AE²=BE²,?

∴x²+(3x)²=1 600,解得x²=160.?

代入AB²=PB·BD,得BD=32.