Question

如图2-26,⊙O的半径为r,MN切⊙O于点A,弦BC交OA于点Q,BP⊥BC,交MN于点P,

求证:(1)PQ∥AC;

图2-26

(2)若AQ=a,AC=b,则.

Answer 1

思路分析:(1)连结AB,应用弦切角定理得∠CAN =∠ABC,运用圆内接四边形的判定和性质说明∠QPA =∠ABC,从而得出结论;(2)过点A作直径AE,连结CE,证△PAQ∽△ECA就可达到目的.

证明:(1)连结AB,∵MN切⊙O于点A,∴OA⊥MN.?

又∵BP⊥BC,∴BA、Q四点共圆,∠QPA =∠ABC.?

又∵∠CAN =∠ABC,∴∠CAN =∠QPA.?

∴PQ∥AC.?

(2)过点A作直径AE,连结CE,则△ECA为直角三角形.?

∵∠CAN =∠E,∠CAN =∠QPA,?

∴∠E =∠QPA.∴Rt△PAQ∽Rt△ECA.?

∴=,  =.