证明:
连接AC
∵AB=CD,AD=BC,AC=CA
∴△ABC≌△CDA(SSS)
∴∠B=∠D
∠BAC=∠DCA=>AB//DC
∠ACB=∠CAD=>AD//BC
∴∠B+∠BAD=180°
∠D+∠BCD=180°
∴∠BAD=∠BCD即∠A=∠C
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