Let ABCD be the cyclic trapezium with AB IICD.
thru' C draw CE parallel to  AD meeting AB in E
.SoAECD is a parallelogram.
angle D=angle AEC.... opp angles of a parallelogram are equal....(i)
angle D+angle ABC=180... opp angles of a cyclic quadr are supplementary....(ii
from (i) and (ii)
angle AEC+angle ABC=180
but triangle AEC+angle CEB= 180...linear pairso angle
ABC= angle CEB ..(iii)
so CE=CB...  sides opp equal angles are equal.(iv) 
butCE=AD...opp  sides of parallelogram AECD.from (iv)
we get
Thus cyclic quadri ABCD is isoceles.
this proves the first part of the question.
now, join AC and BD, the diagonals
.in triangles DAB and CBA,AD=CB...proved before
AB=AB commonangle ADB= angle ACB..
 angles in the same  segment of a circle are equal
.here AB is the triangles DAB and CBA are congruent....SAS rule.
so AD=CB... CPCT 
hence proved.
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