已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC
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已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC
![](http://img.wesiedu.com/upload/4/62/4626b70248393f0f703543e4f45ba478.jpg)
![](http://img.wesiedu.com/upload/4/62/4626b70248393f0f703543e4f45ba478.jpg)
![已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC](/uploads/image/z/4239130-58-0.jpg?t=%E5%B7%B2%E7%9F%A5%3A%E5%A6%82%E5%9B%BE.AB%3DAC%2CAE%3DAD%2CBD%3DEC%2C%E6%B1%82%E8%AF%81%3A%E2%88%A0BAE%3D%E2%88%A0DAC)
∵BD=EC
即BE+ED=ED+DC
∴BE=DC
∵AC=AB、AE=AD
∴△AEB≌△ADC(SSS)
∴∠BAE=∠CAD
即∠BAE=∠DAC
即BE+ED=ED+DC
∴BE=DC
∵AC=AB、AE=AD
∴△AEB≌△ADC(SSS)
∴∠BAE=∠CAD
即∠BAE=∠DAC
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