Geometry

We see that since BDC is an isosceles triangle, so, therefore, angle BDC = angle BCD = x.

38 + 2x = 180

2x = 142

x = 71

From here, we see that angle EDB = 180 - 71 = 109
 

Used as reference:

https://web2.0calc.com/questions/help-asap-with-geometry

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, $\angle A = 90 + 19 = \color{brown}\boxed{109}$