In the diagram below, quadrilateral BAED is a parallelogram, and BD = BC. If angle DBC = \(38^\circ\), then find angle EAB, in degrees.
+580 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
+2668 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, \(\angle A = 90 + 19 = \color{brown}\boxed{109}\)
