题目内容
(2005•太原)如图,直线y=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_ST/0.png)
(1)求证:BD=AO;
(2)在坐标轴上求点E,使得△ODE与△OAB相似;
(3)设点A′在OAB上由O向B移动,但不与点O、B重合,记△OA′B的内心为I,点I随点A′的移动所经过的路程为l,求l的取值范围.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_ST/images1.png)
【答案】分析:(1)利用直线y=
x+2与y轴交于点A,与x轴交于点B,求出A(0,2),B(-2
,0),利用勾股定理求出三角形ABO的边,由边的长度,可求出∠ABO=30°,∠BAO=60°,利用∠BAO的平分线交⊙C于点D,可求出∠ABO=30°=∠BAD,所以BD=AO;
(2)分两种情况:①当∠ODE=90°时,点E的坐标为E1(0,-4),E2(-
,0);
②当∠OED=90°时,E3(0,-1),E4(-
,0);
(3)可设I为△OA'B的内心连接BI,利用动点I到定点D的距离为2,即点I的轨迹是以点D为圆心,2为半径的弧OIB(不含点O、B),可求出弧OIB的长为
,进而求出l的取值范围.
解答:
(1)证明:∵直线y=
x+2与y轴交于点A,与x轴交于点B
∴A(0,2),B(-2
,0),
∴OA=2,0B=2
,AB=4,
∴∠ABO=30°,∠BAO=60°,
∵∠BAO的平分线交⊙C于点D,
∴∠ABO=30°=∠BAD,
∴BD=AO;
(2)解:
①当∠ODE=90°时,点E的坐标为E1(0,-4),E2(-
,0);
②当∠OED=90°时,E3(0,-1),E4(-
,0);
∴符合点E的坐标有四个;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/images11.png)
(3)解:
如图,设I为△OA'B的内心连接BI,连接BH,
∴∠A′BI=∠IBO,
∵BD=OD,∴∠BA′D=∠DBO,
∴∠A′BI+∠BA′D=∠IBO+∠OBD,即∠BID=∠IBD,
∴ID=BD,
∵BD=OA=2,∴ID=2,
∴动点I到定点D的距离为2,即点I的轨迹是以点D为圆心,2为半径的弧OIB(不含点O、B),
弧OIB的长为
,
则l的取值范围是0<l<
.
点评:本题需仔细分析题意,结合图形,利用勾股定理和圆的性质即可解决问题.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/1.png)
(2)分两种情况:①当∠ODE=90°时,点E的坐标为E1(0,-4),E2(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/2.png)
②当∠OED=90°时,E3(0,-1),E4(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/3.png)
(3)可设I为△OA'B的内心连接BI,利用动点I到定点D的距离为2,即点I的轨迹是以点D为圆心,2为半径的弧OIB(不含点O、B),可求出弧OIB的长为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/4.png)
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/images5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/5.png)
∴A(0,2),B(-2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/6.png)
∴OA=2,0B=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/7.png)
∴∠ABO=30°,∠BAO=60°,
∵∠BAO的平分线交⊙C于点D,
∴∠ABO=30°=∠BAD,
∴BD=AO;
(2)解:
①当∠ODE=90°时,点E的坐标为E1(0,-4),E2(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/8.png)
②当∠OED=90°时,E3(0,-1),E4(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/9.png)
∴符合点E的坐标有四个;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/images11.png)
(3)解:
如图,设I为△OA'B的内心连接BI,连接BH,
∴∠A′BI=∠IBO,
∵BD=OD,∴∠BA′D=∠DBO,
∴∠A′BI+∠BA′D=∠IBO+∠OBD,即∠BID=∠IBD,
∴ID=BD,
∵BD=OA=2,∴ID=2,
∴动点I到定点D的距离为2,即点I的轨迹是以点D为圆心,2为半径的弧OIB(不含点O、B),
弧OIB的长为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/10.png)
则l的取值范围是0<l<
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016010_DA/11.png)
点评:本题需仔细分析题意,结合图形,利用勾股定理和圆的性质即可解决问题.
![](http://thumb2018.1010pic.com/images/loading.gif)
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