Question

如图2-14,已知O是正方形ABCD中边BC的中点,AP与以O为圆心,OB为半径的半圆切于T点.求AT∶TP的值.

图2-14

Answer 1

思路分析:注意到AB、AT为切线,PT、PC为切线,则想到连结OA、OT、OP,构造切线长定理的基本图形,要求AT∶TP,则只需求AB∶PC,这可以通过解直角三角形或△ABO∽△OCP求得.

解法一:连结AO、TO、OP.?

∵四边形ABCD为正方形,?

∴BC⊥AB,BC⊥CD.?

又∵BC为⊙O的直径,?

∴AB、DC为⊙O的切线,切点为B、C.?

∵AT、AB切⊙O于T、B,?

∴AT =AB且∠AOB =∠AOT.?

∵PT、PC切⊙O于T、C,?

∴PT =PC且∠POT =∠POC.?

又∵∠AOB +∠AOT +∠POT +∠POC =180°,?

∴∠AOB +∠POC =∠AOP =90°.?

又∠ABO =90°,∴∠POC=∠BAO.?

∴Rt△ABO∽△Rt△OCP.∴= =.?

∴OB =2CP.∴AB =2OC =2OB =4CP,?

即AT∶TP =4∶1.

解法二:先证得∠BAO =∠POC(方法同上).?

在Rt△ABO中,tan∠BAO = =,?

在Rt△OCP中,PC =OC·tan∠POC ==×=,?

∴AT∶TP =4∶1.

解法三:先证得AT =AB,PT =PC(方法同上).?

设正方形边长为a,PT =PC =x,则PD =a-x.?

又∵AT =AB =AD =a,在Rt△ADP中,AD²+DP² =AP²,?

即a²+(a -x)²=(a +x)²,解得.?

∴AT∶TP =4∶1.