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In the diagram below, quadrilateral BAED is a parallelogram, and BD = BC. If angle DBC = \(38^\circ\), then find angle EAB, in degrees.

 

 Apr 14, 2022
 #1
avatar+122 
-2

Hang tight im going to solve it

 Apr 14, 2022
 #2
avatar+326 
+2

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

 Apr 14, 2022
 #3
avatar+122 
-1

I believe it equal 109 97%sure

 Apr 14, 2022
 #4
avatar+1384 
+1

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

 Apr 14, 2022
 #5
avatar+122390 
0

Nice solutions ,guys  !!!!

 

 

cool cool cool

CPhill  Apr 14, 2022

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