I think the answer is NO,but it should be a isosceles trapezoid and $AD=DC=CB$.Proof like this. Let $AC$ meets $BD$ at $M$. Due to $\angle DBC=\angle CAB=\angle DCA$,we have $DC^2=DM\cdot DB$,recall $\angle DAC=\angle ABD$,we have $DA^2=DM\cdot DB$. So $DA=DC$,simillarly $CD=CB$.$\Box$

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SBYT wrote: I think the answer is NO,but it should be a isosceles trapezoid and $AD=DC=CB$.Proof like this. Let $AC$ meets $BD$ at $M$. Due to $\angle DBC=\angle CAB=\angle DCA$,we have $DC^2=DM\cdot DB$,recall $\angle DAC=\angle ABD$,we have $DA^2=DM\cdot DB$. So $DA=DC$,simillarly $CD=CB$.$\Box$ how did you conclude $DC^2=DM \cdot DB$