Here's another way to solve it. Because \(\triangle{BCD}\) is isosceles, \(\angle D = \angle C = 71 \)
Because \(BAED\) is a parallelogram, \(\angle D = \angle E = 71\).
Extend Point A to M, so it is perpendicular to \(\overline {EC}\) .
\(\angle A = \angle{EAM} + \angle {MAB}\)
In triangle \(\triangle {AME} \), \(\angle EAM = 19\).
Because \(\overline {AM}\) is perpendicular to \(\overline{BA}\), \(\angle{MAB} = 90\)
Thus, \(\angle A = 90 + 19 = \color{brown}\boxed{109}\)
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i couldent help 
