Let ABCD be the cyclic quadrilateral with AD parallel to BC. The center of the circle is O.
Now, since AD is parallel to BC and AC is the transverse
angle DAC = angle ACB (interior opposite angles)
angle DAC is subtended by chord DC on the circumference and angle ACB is subtended by chord AB.
If the angles subtended by 2 chords on the circumference of the circle are equal, then the lengths of the chords is also equal.
Hence, DC = AB i.e. the 2 remaining sides of the cyclic quadrilateral are equal.
Now, BD and AC are the 2 diagonals.
In triangle ABD and triangle DCA
AB = CD (proved above)
angle DCA = angle DAB (angles subtended by the same chord AD)
angle ACB = angle DAC(proved above)
Hence, triangle ABD is congruent to triangle DCA (by ASA)
hence, BD = AC (by CPCT)