Draw this cyclic quadilateral and mark ABCD in order.

Draw the two diagonals (BC and AD)

Now in ADC Triangle.

DAC angle = DCA angle (since AD = DC)

Let's call this two angles as x,

therefore, DCA angle = DAC angle = x

Therefore ADC angle = 180 - 2x (Angles in a triangle)

But in a cyclic quadilateral the opposing angles are complementary,

ADC angle + ABC angle = 180

180 -2x + ABC angle = 180

ABC angle = 2x

Conside the circle segment AD,

Now DCA angle = DBA angle (Angles in same segment)

Since DCA angle = x

Therefore DBA angle = x

But

ABC angle = DBA angle + DBC angle

2x = x + DBC angle

DBC angle = x

but in DBC trinagle, DC = BC

Therefore,

DBC angle = CDB angle = x

Therefore in DBC triangle ,

DCB angle = 180 -2x

Now,

DCB angle = 180 -2x

and

ADC angle = 180 -2x (previously proved)

Therefore ADC angle = DCB angle

CDB angle = x (previous proved)

and

DBA angle  = x (previously proved)

Therefore DC // AB (Properties of parallelogram)