题目内容
![](http://thumb.1010pic.com/pic3/upload/images/201201/88/29b56233.png)
1 | 2 |
分析:作一线段BC,使BC=a,然后作BC的垂直平分线,垂足为D,再以点D为圆心,以
a为半径画弧,与a的垂直平分线相交于点A,然后连接AB、AC即可得解.
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2 |
解答:已知:线段a![](http://thumb.1010pic.com/pic3/upload/images/201201/92/54de1252.png)
求作:△ABC,使AB=AC,BC=a,AD⊥BC,且AD=
a.
![](http://thumb.1010pic.com/pic3/upload/images/201201/92/8fab7047.png)
![](http://thumb.1010pic.com/pic3/upload/images/201201/92/54de1252.png)
求作:△ABC,使AB=AC,BC=a,AD⊥BC,且AD=
1 |
2 |
![](http://thumb.1010pic.com/pic3/upload/images/201201/92/8fab7047.png)
点评:本题考查了作一条线段等于已知线段,线段垂直平分线的作法,理清作等腰三角形的思路是解题的关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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