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AB=AB,角A=20度,則角B=80度.以AB為邊長作等邊三角形ABE(點E與C在AB兩側),連接DE.則:AE=AB=AC;且∠DAE=∠DAB+∠BAE=80°=∠C.又AD=BC.故⊿DAE≌ΔBCA(SAS),得:DE=AB;且∠AED=∠CAB=20°.DE=AC=AE=BE,則點A,D,B在以點E…
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