Question

如圖2-4-17,AB是⊙O的直徑,PB切⊙O于點B,PA交⊙O于點C,∠APB的平分線分別交BC、AB于點D、E,交⊙O于點F,∠A=60°,并且線段AE、BD的長是一元二次方程x²-kx +=0的兩個根(k為常數(shù)).

圖2-4-17

(1)求證:PA·BD=PB·AE;

(2)證明⊙O的直徑長為常數(shù);

(3)求tan∠FPA的值.

Answer 1

思路分析:(1)由△PBD∽△PAE即可證得.?

(2)由韋達定理知AE +BD =k,只需證BE =BD,這可由角的相等證得.?

(3)要求tan∠FPA,先將∠FPA轉化到直角三角形中,而∠FPB =∠FPA,∠FPB恰好在Rt△PBE中,解此三角形即可.

(1)證明:∵PB切⊙O于點B,∴∠PBD =∠A.?

又PE平分∠APB,∴∠APE =∠BPD.?

∴△PBD∽△PAE.∴=.?

∴PA·BD = PB·AE.

(2)解:由(1)知∠APE =∠EPB,?

又∵∠BED =∠A +∠EPA,∠BDE =∠PBC+∠EPB,?

∴∠BED =∠BDE.∴BE =BD.?

∵AE、BD為方程x²-kx +=0的兩個根,?

∴AE +BD =k =AB.?

∴⊙O的直徑為常數(shù)k.

(3)解:∵PB切⊙O于點B,AB為直徑,?

∴∠PBA =90°.∵∠A =60°,?

∴PB =PA·sin60°=.?

由(1)得PA·BD =PB·AE,?

∴.?

∵AE、BD的長是方程x²-kx +=0的兩個根,

∴AE·BD =.?

∴AE =2,BD =∴.?

在Rt△PBA中,PB =AB·tan60°=()·=.?

在Rt△PBE中,tan∠BPE = = =,?

又∠FPA =∠BPF,∴tan∠FPA =.