How??
it should be 105o.
taking tri(ABC), given ang(BAC) = 30 and AB = BC, so ang(ACB) = ang(ABC).
sum of all angles of a triangle = 180o
therefore, ang(ABC+ACB+BAC) = 180o and BAC = 30o
=> 2*ang(ACB) = 150o since ACB = ABC
=> ACB = 75o
=> ang(CAD) = ang(ACB) = 75o, property of ||ogram, alternate angles of a diagnol are equal
=> BAD = BAC + CAD = 30o + 75o = 105o
also, BAD = ADE, property of || lines
therefore, ADE = 105o
WAIT, could you please re-read and re-post your question!! if AC = AB then it is no longer a ||gram but a quadrilateral!!