题目内容
如图,射线AM平行于射线BN,AB⊥BN且AB=3,C是射线BN上的一个动点,连接AC,作CD⊥AC且CD=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_ST/0.png)
(1)AC长为______
【答案】分析:(1)由AB⊥BN且AB=3,BC长为t,根据勾股定理的知识,即可求得AC的长,由作CD⊥AC且CD=
AC,根据三角形面的求解方法即可求得△ACD的面积;
(2)过D作DF⊥BN交BN于点F,由∠ABC=∠CFD=90°,∠FDC=∠ACB,即可得△DFC∽△CBA,然后根据相似三角形的对应边成比例,即可求得点D到射线BN的距离;
(3)分别从①当EC=AE时,E为AD中点,EC=
AD,②当AE=AC时,AM⊥DF,③当0≤t<12时,∠AEC为钝角,故AC≠CE,当t≥12时,CE≤DF<DC<AC去分析求解,即可得到当BC等于
和6+3
时,△ACE为等腰三角形.
解答:
解:(1)∵AB⊥BN,
∴∠B=90°,
∵AB=3,BC长为t,
∴AC=
=
;
∵CD=
AC=
,
∵CD⊥AC,
∴∠AD=90°,
∴△ACD的面积为:
AC•CD=
×
×
=
;
(2)过D作DF⊥BN交BN于点F,
∵∠ABC=∠CFD=90°,∠FDC=∠ACB,
∴△DFC∽△CBA.
∴
=
=
,
∴DF=
,BC=t.
即点D到射线BN的距离为
;
(3)①如图,当EC=AE时,E为AD中点,EC=
AD,
此时FC=BC,
∴t=
;
②如图,∵EC⊥BN,
∴AE≠AC,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/images21.png)
③当t=0时,C与B重合,CD=
AC,
可得DF=t=0,此时△AEC不能为等腰直角三角形,
当t=12时,CE≤DF<DC<AC,
∴当0≤t<12时,∠AEC为钝角,故AC≠CE,△ACE不能为等腰三角形;
当t≥12时,CE≤DF<DC<AC,此时△ACE不能为等腰三角形,
综上所述,当BC等于
时,△ACE为等腰三角形.
点评:此题考查了相似三角形的判定与性质,直角三角形的性质等知识.解此题的关键是注意分类讨论思想,数形结合思想与方程思想的应用.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/0.png)
(2)过D作DF⊥BN交BN于点F,由∠ABC=∠CFD=90°,∠FDC=∠ACB,即可得△DFC∽△CBA,然后根据相似三角形的对应边成比例,即可求得点D到射线BN的距离;
(3)分别从①当EC=AE时,E为AD中点,EC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/3.png)
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/images4.png)
∴∠B=90°,
∵AB=3,BC长为t,
∴AC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/5.png)
∵CD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/7.png)
∵CD⊥AC,
∴∠AD=90°,
∴△ACD的面积为:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/12.png)
(2)过D作DF⊥BN交BN于点F,
∵∠ABC=∠CFD=90°,∠FDC=∠ACB,
∴△DFC∽△CBA.
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/15.png)
∴DF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/16.png)
即点D到射线BN的距离为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/17.png)
(3)①如图,当EC=AE时,E为AD中点,EC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/18.png)
此时FC=BC,
∴t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/19.png)
②如图,∵EC⊥BN,
∴AE≠AC,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/images21.png)
③当t=0时,C与B重合,CD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/20.png)
可得DF=t=0,此时△AEC不能为等腰直角三角形,
当t=12时,CE≤DF<DC<AC,
∴当0≤t<12时,∠AEC为钝角,故AC≠CE,△ACE不能为等腰三角形;
当t≥12时,CE≤DF<DC<AC,此时△ACE不能为等腰三角形,
综上所述,当BC等于
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191804697484880/SYS201311011918046974848010_DA/21.png)
点评:此题考查了相似三角形的判定与性质,直角三角形的性质等知识.解此题的关键是注意分类讨论思想,数形结合思想与方程思想的应用.
![](http://thumb2018.1010pic.com/images/loading.gif)
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