题目

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 答案：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.

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