题目内容
(2003•泰州)已知:如图,⊙O与⊙O1内切于点A,AO是⊙O1的直径,⊙O的弦AC交⊙O1于点B,弦DF经过点B且垂直于OC,垂足为点E.(1)求证:DF与⊙O1相切;
(2)求证:2AB2=AD•AF;
(3)若AB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_ST/images2.png)
【答案】分析:(1)本题可连接O1B,证O1B⊥DF即可,由于OC⊥DF,因此只需证O1B∥OC即可.可通过不同圆中圆的半径对应的角相等来求得,由此可得证.
(2)本题可通过证△ABD和△AFC相似来求解.连接OB,则OB⊥AC,因此可根据垂径定理得出AC=2AB,那么通过两三角形相似得出的AD:AC=AB:AF,即可得出所求的结论.
(3)本题可先求出BF的长,然后根据相似三角形FCB和ACF得出的CF 2=CB•CA,求出CF的长,还是这两个相似三角形,根据CF:AF=BC:CF求出AF的长,进而可根据(2)的结果求出AD的长.
解答:
(1)证明:连接O1B,
∵O1B=O1A,
∴∠O1AB=∠O1BA.
∵OA=OC,
∴∠OAC=∠OCA.
∴∠O1BA=∠OCA.
∴O1B∥OC.
∵OC⊥DF,
∴O1B⊥DF.
∴DF与⊙O1相切.
(2)证明:连接OB,则OB⊥AC,
∴AC=2AB=2BC.
∵OC⊥DF,
∴弧DC=弧CF.
∴∠CAD=∠CAF.
∵∠D=∠ACF,
∴△ABD∽△AFC.
∴
.
∵AC=2AB,
∴2AB2=AD•AF.
(3)解:直角△BEC中,BC=AB=2
,cos∠CBE=cos∠DBA=
=
,
∴BE=2,CE=4.
∵直角△OBE中,∠BOE=∠CBE=90°-∠BCO,BE=2,
∴BO=
,OE=1.
∴AO=OC=OE+EC=5.
连接OF,直角△OEF中,OF=OA=5,OE=1,根据勾股定理有EF=2
,
∴BF=2
+2.
∵弧DC=弧CF,
∴∠CAF=∠BFC.
∴△ACF∽△FCB.
∴CF2=CB•CA=2AB2=40.
∴CF=2
.
∴
.
即
=
,
∴AF=4
+2
.
由(2)知:2AB2=AD•AF.
∴AD=4
-2
.
点评:本题主要考查了切线的判定、垂径定理、相似三角形的判定和性质等知识点,在(3)中通过相似三角形求出CF、AF的长是解题的关键.
(2)本题可通过证△ABD和△AFC相似来求解.连接OB,则OB⊥AC,因此可根据垂径定理得出AC=2AB,那么通过两三角形相似得出的AD:AC=AB:AF,即可得出所求的结论.
(3)本题可先求出BF的长,然后根据相似三角形FCB和ACF得出的CF 2=CB•CA,求出CF的长,还是这两个相似三角形,根据CF:AF=BC:CF求出AF的长,进而可根据(2)的结果求出AD的长.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/images0.png)
∵O1B=O1A,
∴∠O1AB=∠O1BA.
∵OA=OC,
∴∠OAC=∠OCA.
∴∠O1BA=∠OCA.
∴O1B∥OC.
∵OC⊥DF,
∴O1B⊥DF.
∴DF与⊙O1相切.
(2)证明:连接OB,则OB⊥AC,
∴AC=2AB=2BC.
∵OC⊥DF,
∴弧DC=弧CF.
∴∠CAD=∠CAF.
∵∠D=∠ACF,
∴△ABD∽△AFC.
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/0.png)
∵AC=2AB,
∴2AB2=AD•AF.
(3)解:直角△BEC中,BC=AB=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/3.png)
∴BE=2,CE=4.
∵直角△OBE中,∠BOE=∠CBE=90°-∠BCO,BE=2,
∴BO=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/4.png)
∴AO=OC=OE+EC=5.
连接OF,直角△OEF中,OF=OA=5,OE=1,根据勾股定理有EF=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/5.png)
∴BF=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/6.png)
∵弧DC=弧CF,
∴∠CAF=∠BFC.
∴△ACF∽△FCB.
∴CF2=CB•CA=2AB2=40.
∴CF=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/7.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/8.png)
即
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/10.png)
∴AF=4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/12.png)
由(2)知:2AB2=AD•AF.
∴AD=4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232426423390766/SYS201310212324264233907025_DA/14.png)
点评:本题主要考查了切线的判定、垂径定理、相似三角形的判定和性质等知识点,在(3)中通过相似三角形求出CF、AF的长是解题的关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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