Given that \(AB=DC,DA \)perpendicular to \(AB\) .
Let M and N be the perpendicular bisector of \(AD\) and \(BC\) respectively.
These two perpendicular bisector intersect in \(E\).
since \(EM\) is perpendicular bisector \(AE=ED\)
since \(EN\) is perpendicular bisector \(EB=EC\)
\(\triangle EAB=\triangle EDC\) (SSS)
\(\angle EAB=\angle EDC\)
\(\angle EAD=\angle EDA\)
we have \(\angle MAB=\angle MDC\)
obviously it is impossible.