题目内容
如图,以等腰直角三角形ABC的斜边AB为边向内作等边△ABD,连接DC,以DC为边作等边△DCE.B、E在C、D的同侧,若AB=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_ST/images1.png)
【答案】分析:由等腰直角三角形ABC中,AB=
,由勾股定理可知AC=
AB=1,再证△ADC≌△BDE,从而推出BE=AC=1.
解答:解:∵等腰直角三角形ABC中,AB=
,
∴AC=
AB=1,
∵等边△ABD和等边△DCE,
∴AD=BD,CD=ED,∠ADB=∠CDE,
∴∠ADC=∠BDE,
在△ADC和△BDE中,
,
∴△ADC≌△BDE(SAS),
∴BE=AC=1.
点评:解决本题的关键是利用三角形全等得到所求线段的转化.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_DA/1.png)
解答:解:∵等腰直角三角形ABC中,AB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_DA/2.png)
∴AC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_DA/3.png)
∵等边△ABD和等边△DCE,
∴AD=BD,CD=ED,∠ADB=∠CDE,
∴∠ADC=∠BDE,
在△ADC和△BDE中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202046140517211/SYS201311032020461405172004_DA/4.png)
∴△ADC≌△BDE(SAS),
∴BE=AC=1.
点评:解决本题的关键是利用三角形全等得到所求线段的转化.
![](http://thumb2018.1010pic.com/images/loading.gif)
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