Construct lines AE, CD, BC (extend BC to X)
Let the intersection of AE and BC be F, < ABC = a, < CBE = b
So < ABC = < ACB = a (equal <'s opp equal sides)
But < ACB = < AEB = a (<'s subtended by same arc)
Considering ∆ BEF, < BFE = 180 - (a + b) (< sum of ∆)
So < AFC = 180 - (a + b) (vert opp <'s)
But < BAD = 180 - (a + b) (co-int <'s, AD||BE)
So < BAD = < DCX = 180 - (a + b) (ext < of cyclic quad ABCD)
ie. < AFC = < DCX
ie. AE||DC (corresp <'s)
If I had a diagram that would make it much easier :|