已知:如图,AE=AC,AD= AB,∠EAC=∠DAB,求证:△EAD全等于△CAB
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已知:如图,AE=AC,AD=AB,∠EAC=∠DAB,求证:△EAD全等于△CAB已知:如图,AE=AC,AD=AB,∠EAC=∠DAB,求证:△EAD全等于△CAB已知:如图,AE=AC,AD=A
已知:如图,AE=AC,AD= AB,∠EAC=∠DAB,求证:△EAD全等于△CAB
已知:如图,AE=AC,AD= AB,∠EAC=∠DAB,求证:△EAD全等于△CAB
已知:如图,AE=AC,AD= AB,∠EAC=∠DAB,求证:△EAD全等于△CAB
证明:
∵∠BAC=∠DAB+∠DAC,∠DAE=∠EAC+∠DAC,∠EAC=∠DAB
∴∠BAC=∠DAE
∵AE=AC,AD=AB
∴△EAD≌△CAB (SAS)
因为∠EAC=∠DAB,所以∠EAD=∠CAB;
又AE=AC,AD= AB;
所以△EAD全等于△CAB(边角边)
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