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In the diagram below, quadrilateral $ABDE$ is a parallelogram, and $BC = BD$. If $\angle CBD = 28^\circ$, then find $\angle BAE$, in degrees.

eileenthecoolbean  Aug 7, 2017
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I found this diagram from here: https://www.mathisfunforum.com/viewtopic.php?id=22434

 

 

There are  180°  in every triangle.  So..

 

m∠BDC+m∠BCD+ m∠CBD=180°  
                                      The problem says that  m∠CBD  =  28°
m∠BDC+m∠BCD+28°=180°  
        Since  BD  =  BC,  m∠BCD  =  m∠BDC
m∠BDC+m∠BDC+28°=180°  
         
2(m∠BDC)+28°=180°  
        Subtract  28°  from both sides.
2(m∠BDC)  =152°  

 

       

Divide both sides of the equation by  2  .

  m∠BDC  =76°  

 

 

And since  m∠BDC  and  m∠BDE  form a straight line, they sum to  180° .

 

76°  +  m∠BDE  =  180°

                                              Subtract  76°  from both sides.

            m∠BDE  =  104°

 

 

In a parallelogram, opposite angles are congruent. Imagine sliding up ED onto AB and sliding over BD onto AE. Then you can see that they are vertical angles, so they have the same measure.

 

m∠BDE  =  m∠BAE  =  104°

hectictar  Aug 7, 2017
edited by hectictar  Aug 7, 2017
edited by hectictar  Aug 7, 2017

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