题目内容
(2009•深圳)如图,AB是⊙O的直径,AB=10,DC切⊙O于点C,AD⊥DC,垂足为D,AD交⊙O于点E.(1)求证:AC平分∠BAD;
(2)若sin∠BEC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_ST/images1.png)
【答案】分析:(1)连接OC,易证AD∥OC,则∠DAC=∠ACO,则只要证明∠CAO=∠ACO,根据等边对等角即可证明;
(2)∠BEC=∠BAC,则直角△ABC中即可求得∠ABC,根据三角函数即可求得AB、AC的长,而∠DCA=∠CBA,在直角△ACD中即可利用三角函数求得CD的长.
解答:
(1)证明:连接OC,由DC是切线得OC⊥DC;
又AD⊥DC,
∴AD∥OC,
∴∠DAC=∠ACO.
又由OA=OC得∠BAC=∠ACO,
∴∠DAC=∠BAC.
即AC平分∠BAD.
(2)解:方法一:∵AB为直径,
∴∠ACB=90°
又∵∠BAC=∠BEC,
∴BC=AB•sin∠BAC=AB•sin∠BEC=6.
∴AC=
.
又∵∠DAC=∠BAC=∠BEC,且AD⊥DC,
∴CD=AC•sin∠DAC=AC•sin∠BEC=
.
方法二:∵AB为直径,
∴∠ACB=90°.
又∵∠BAC=∠BEC,
∴BC=AB•sin∠BAC=AB•sin∠BEC=6.
∴
.
又∵∠DAC=∠BAC,∠D=∠ACB=90°,
∴△ADC∽△ACB,
,即
,
解得
.
点评:本题考查了圆的切线的性质及解直角三角形的知识.运用切线的性质来进行计算或论证,常通过作辅助线连接圆心和切点,利用垂直构造直角三角形解决有关问题.
(2)∠BEC=∠BAC,则直角△ABC中即可求得∠ABC,根据三角函数即可求得AB、AC的长,而∠DCA=∠CBA,在直角△ACD中即可利用三角函数求得CD的长.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/images0.png)
又AD⊥DC,
∴AD∥OC,
∴∠DAC=∠ACO.
又由OA=OC得∠BAC=∠ACO,
∴∠DAC=∠BAC.
即AC平分∠BAD.
(2)解:方法一:∵AB为直径,
∴∠ACB=90°
又∵∠BAC=∠BEC,
∴BC=AB•sin∠BAC=AB•sin∠BEC=6.
∴AC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/0.png)
又∵∠DAC=∠BAC=∠BEC,且AD⊥DC,
∴CD=AC•sin∠DAC=AC•sin∠BEC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/1.png)
方法二:∵AB为直径,
∴∠ACB=90°.
又∵∠BAC=∠BEC,
∴BC=AB•sin∠BAC=AB•sin∠BEC=6.
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/2.png)
又∵∠DAC=∠BAC,∠D=∠ACB=90°,
∴△ADC∽△ACB,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/4.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021225837257914132/SYS201310212258372579141023_DA/5.png)
点评:本题考查了圆的切线的性质及解直角三角形的知识.运用切线的性质来进行计算或论证,常通过作辅助线连接圆心和切点,利用垂直构造直角三角形解决有关问题.
![](http://thumb2018.1010pic.com/images/loading.gif)
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