题目内容
![](http://thumb.1010pic.com/pic3/upload/images/201010/22/1c80b9cf.png)
分析:在DC上取DE=DB.连接AE,在Rt△ABD和Rt△AED中,BD=ED,AD=AD.证明△ABD≌△AED即可求解;
解答:解:如图,
![](http://thumb.1010pic.com/pic3/upload/images/201103/6/c1bea4ea.png)
在DC上取DE=DB.连接AE,在Rt△ABD和Rt△AED中,BD=ED,AD=AD.
∴△ABD≌△AED.
∴AB=AE,∠B=∠AED.
又∵AB+BD=CD
∴EC=CD-DE=CD-BD=(AB+BD)-BD=AB=AE
∴∠C=∠CAE
∴∠B=∠AED=2∠C
又∵∠B+∠C=180°-∠BAC=60°
∴∠C=20°,
故选A.
![](http://thumb.1010pic.com/pic3/upload/images/201103/6/c1bea4ea.png)
在DC上取DE=DB.连接AE,在Rt△ABD和Rt△AED中,BD=ED,AD=AD.
∴△ABD≌△AED.
∴AB=AE,∠B=∠AED.
又∵AB+BD=CD
∴EC=CD-DE=CD-BD=(AB+BD)-BD=AB=AE
∴∠C=∠CAE
∴∠B=∠AED=2∠C
又∵∠B+∠C=180°-∠BAC=60°
∴∠C=20°,
故选A.
点评:本题考查了等边三角形的判定与性质及三角形内角和定理,属于基础图,关键是巧妙作出辅助线.
![](http://thumb2018.1010pic.com/images/loading.gif)
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