如图,已知AB=AE,∠BAE=∠CAD,AC=AD,求证:(1)△ABC≌△AED; (2)BC=ED.
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![如图,已知AB=AE,∠BAE=∠CAD,AC=AD,求证:(1)△ABC≌△AED; (2)BC=ED.](/uploads/image/z/5200556-68-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5AB%3DAE%2C%E2%88%A0BAE%3D%E2%88%A0CAD%2CAC%3DAD%2C%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%89%E2%96%B3ABC%E2%89%8C%E2%96%B3AED%EF%BC%9B++++++++++++++++++++++%EF%BC%882%EF%BC%89BC%3DED.)
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如图,已知AB=AE,∠BAE=∠CAD,AC=AD,求证:(1)△ABC≌△AED; (2)BC=ED.
如图,已知AB=AE,∠BAE=∠CAD,AC=AD,求证:
(1)△ABC≌△AED; (2)BC=ED.
如图,已知AB=AE,∠BAE=∠CAD,AC=AD,求证:(1)△ABC≌△AED; (2)BC=ED.
这题太简单了吧?
∵∠BAE=∠CAD,
∴∠BAE+∠EAC=∠CAD+∠EAC
即:∠BAC=∠EAD.
在△ABC与△AED中
∵ AB=AE, ∠BAC=∠EAD,AC=AD ,
∴△ABC≌△AED (边角边定理)
∴BC=ED (全等三角形对应边相等)
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