题目内容
已知AC⊥BC于C,BC=a,CA=b,AB=c,下列选项中⊙O的半径为
的是( )
ab |
a+b |
A.![]() | B.![]() | C.![]() | D.![]() |
A、设圆的半径是x,圆切AC于E,切BC于D,切AB于F,如图(1)同样得到正方形OECD,AE=AF,BD=BF,则a-x+b-x=c,求出x=
,故本选项错误;
B、设圆切AB于F,圆的半径是y,连接OF,如图(2),
则△BCA∽△OFA,∴
=
,
∴
=
,解得:y=
,故本选项错误;
C、连接OE、OD,
∵AC、BC分别切圆O于E、D,
∴∠OEC=∠ODC=∠C=90°,
∵OE=OD,
∴四边形OECD是正方形,
∴OE=EC=CD=OD,
设圆O的半径是r,
∵OE∥BC,∴∠AOE=∠B,
∵∠AEO=∠ODB,
∴△ODB∽△AEO,
∴
=
,
=
,
解得:r=
,故本选项正确;
D、O点连接三个切点,从上至下一次为:OD,OE,OF;并设圆的半径为x;
容易知道BD=BF,所以AD=BD-BA=BF-BA=a+x-c;
又∵b-x=AE=AD=a+x-c;所以x=
,故本选项错误.
故选C.
a+b-c |
2 |

B、设圆切AB于F,圆的半径是y,连接OF,如图(2),
则△BCA∽△OFA,∴
OF |
BC |
AO |
AB |
∴
y |
a |
b-y |
c |
ab |
a+c |

C、连接OE、OD,
∵AC、BC分别切圆O于E、D,
∴∠OEC=∠ODC=∠C=90°,
∵OE=OD,
∴四边形OECD是正方形,
∴OE=EC=CD=OD,

设圆O的半径是r,
∵OE∥BC,∴∠AOE=∠B,
∵∠AEO=∠ODB,
∴△ODB∽△AEO,
∴
OE |
BD |
AE |
OD |
r |
a-r |
b-r |
r |
解得:r=
ab |
a+b |

D、O点连接三个切点,从上至下一次为:OD,OE,OF;并设圆的半径为x;
容易知道BD=BF,所以AD=BD-BA=BF-BA=a+x-c;
又∵b-x=AE=AD=a+x-c;所以x=
b+c-a |
2 |
故选C.

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