证明:
设四边形ABCD,
AC平分对角即∠1=∠2,∠5=∠6,
BD平分对角即∠3=∠4,∠7=∠8
∵∠AOB=∠COD
∴∠2+∠3=∠6+∠7
∴2∠2+2∠3=2∠6+2∠7
即∠DAB+∠ABC=∠BCD+∠ADC
∵∠DAB+∠ABC+∠BCD+∠ADC=360°
∴∠DAB+∠ABC=∠BCD+∠ADC=180°
∴AD//BC
∴∠1=∠5,∠8=∠4
∴∠1=∠6=>AD=CD
∠8=∠3=>AD=AB
∠7=∠4=>CD=BC
∴AB=BC=CD=AD
∴四边形ABCD是菱形
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