Question

如圖2-5-19,已知PA為⊙O的切線(xiàn),PO交⊙O于點(diǎn)B,BC⊥PA于點(diǎn)C,交⊙O于點(diǎn)D,

圖2-5-19

(1)求證:AB²=PB·BD.

(2)若PA =15,PB =5,求BD的長(zhǎng).

Answer 1

思路分析:(1)只需證△PBA∽△ABD.?

(2)在(1)的基礎(chǔ)上,只需求AB,因此尋找AB與BE的關(guān)系式,這可以通過(guò)相似三角形和勾股定理達(dá)到目的.

(1)證明:連結(jié)AD,延長(zhǎng)PO交⊙O于E,連結(jié)AE.?

∵BC⊥PA,∴∠P +∠PBC =90°.?

∵BE為直徑,?

∴∠BAE =90°,∠BAD +DAE =90°.?

∵∠DAE =∠DBE =∠PBC,∴∠P =∠BAD.?

又∵∠PAB =∠ADB,∴△PBA∽△ABD.?

∴=,即AB² =PB·BD.

(2)解:∵PA為切線(xiàn),∴PA²=PB·PE.?

又PA =15,PB =5,∴PE =45.?

∴BE =40.?

∵△PBA∽△PAE,∴= ==.?

設(shè)AB =x,則AE =3x.

又AB²+AE²=BE²,?

∴x²+(3x)²=1 600,解得x²=160.?

代入AB²=PB·BD,得BD=32.