题目内容
如图,D为△ABC的AB边上的一点,∠DCA=∠B,若AC=![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_ST/images1.png)
A.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_ST/1.png)
B.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_ST/2.png)
C.2cm
D.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_ST/3.png)
【答案】分析:先判断△ADC与△ACB相似,再利用相似三角形对应边成比例求解即可.
解答:解:∵∠A=∠A,∠DCA=∠B,
∴△ADC∽△ACB,
∴AD:AC=AC:AB,
∵AC=
cm,AB=3cm,
∴AD:
=
:3,
解得AD=2cm.
故选C.
点评:此题主要考查相似三角形的判定及性质.
解答:解:∵∠A=∠A,∠DCA=∠B,
∴△ADC∽△ACB,
∴AD:AC=AC:AB,
∵AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_DA/0.png)
∴AD:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103943183586506/SYS201312111039431835865012_DA/2.png)
解得AD=2cm.
故选C.
点评:此题主要考查相似三角形的判定及性质.
![](http://thumb.zyjl.cn/images/loading.gif)
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