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Can anyone provide the answer for this?  TheXSquaredFactor was very helpful but I think it would make more sense if I could see the final product.

 

Thanks!

AnonymousConfusedGuy  Jan 23, 2018
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3+0 Answers

 #1
avatar+85759 
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What's the question  ???

 

 

cool cool cool

CPhill  Jan 23, 2018
 #2
avatar+925 
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A quadrilateral is called a "parallelogram" if both pairs of opposite sides are parallel. Show that if WXYZ is a parallelogram, then < W = < Y and < X = < Z.  

 

(< = angle)

 

I put the link in the problem title.

AnonymousConfusedGuy  Jan 23, 2018
 #3
avatar+1871 
+1

Go here https://web2.0calc.com/questions/not-sure-where-to-start-here to view my original response and the corresponding question. 

 

I already explained in the previous post that opposite sides of a parallelogram are parallel by definition. Then, I encouraged you to try and figure out a relationship with two angles located on the same segment. In this case, \(\angle Z\) and \(\angle Y\) are same-side interior angles (sometimes referred to as consecutive interior angles). I know this because \(\overline{WZ}\parallel\overline{XY}\), and \(\overline{ZY}\) acts as the transversal.  Because there is proof of parallel segments, we also know that the same-side interior angles are supplementary. If the angles are known to be supplementary, then the sum of the angles equals 180°. By the definition of supplementary angles, then, \(m\angle Z+M\angle Y=180^{\circ}\). Using the same logic as before, it is also possible to conclude that \(m\angle X+m\angle Y=180^{\circ}\). Based on the two previous conclusions, both \(\angle Z\) and \(\angle X\) are supplementary to a common angle, \(\angle Y\) in this case. When this occurs, the congruent supplements theorem states that \(\angle Z\cong\angle X\). Utilize this logic again to prove that the other pair of opposite angles, \(\angle W\text{ and }\angle Y\), are indeed congruent. The process is identical. 

TheXSquaredFactor  Jan 23, 2018

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