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.
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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)