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# Geometry

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167
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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
+115
-2

Hang tight im going to solve it

Apr 14, 2022
#2
+579
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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
+115
-1

I believe it equal 109 97%sure

Apr 14, 2022
#4
+2541
+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
+124696
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Nice solutions ,guys  !!!!

CPhill  Apr 14, 2022